Reference Sheet · Plane

Line through two points
Two-point form
\[\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}\]
Slope
\[k=\frac{y_2-y_1}{x_2-x_1}\]
General form \(Ax+By+C=0\)
Slope from general form
\[k=-\frac{A}{B}\]
Distance from point \((x_0,y_0)\)
\[d=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\]
Midpoint
\[E=\left(\frac{x_1+x_2}{2},\;\frac{y_1+y_2}{2}\right)\]
Angle between lines
Both slopes finite
\[\tan\varphi=\left|\frac{k_1-k_2}{1+k_1k_2}\right|\]
One line vertical \((k\to\infty)\)
\[\tan\varphi=\left|\frac{1}{k}\right|\]
Parallel
\(k_1=k_2\;\Rightarrow\;\varphi=0°\)
Perpendicular
\(k_1\cdot k_2=-1\;\Rightarrow\;\varphi=90°\)
Equidistant lines through a point
Either parallel to \(BC\), or through midpoint of \(BC\)

Pick a section, review the key definitions, then work through each problem one guided step at a time — with hints, instant feedback, and diagrams that build as you solve.

Analytic Geometry in the Plane

Lines in the coordinate plane — equations, slopes, distances and angles.

Analytic Geometry in Space

Planes and lines in three dimensions — equations, distances and angles.

Vector Algebra

Vectors given by coordinates — lengths, angles, and the scalar, vector and mixed products.

Curves of the Second Order

Ellipse, hyperbola and parabola — type from A·C, canonical form, foci, eccentricity, directrices.
More sections are on the way — check back regularly for new topics.
Analytic Geometry in the Plane

Basic Definitions

The core formulas for straight lines in the coordinate plane — the tools used throughout the problems below.

I.Division of a Segment in a Given Ratio

If two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) are given, the point \(C(x,y)\) on the same line dividing the segment in the ratio \(\lambda\) has coordinates:

\[x=\frac{x_1+\lambda x_2}{1+\lambda},\qquad y=\frac{y_1+\lambda y_2}{1+\lambda}\]

II.General Equation of a Line

Every equation of the first degree in \(x\) and \(y\), i.e. an equation of the form \(Ax+By+C=0\) where \(A,B,C\) are constants, defines a certain line in the plane. This is called the general equation of a line. Particular cases:

  • \(C=0,\ A\neq0,\ B\neq0\): the line \(Ax+By=0\) passes through the origin.
  • \(A=0,\ B\neq0,\ C\neq0\): the line \(By+C=0\) is parallel to the \(Ox\) axis.
  • \(B=0,\ A\neq0,\ C\neq0\): the line \(Ax+C=0\) is parallel to the \(Oy\) axis.
  • \(B=C=0,\ A\neq0\): the line \(Ax=0\) coincides with the \(Oy\) axis.
  • \(A=C=0,\ B\neq0\): the line \(By=0\) coincides with the \(Ox\) axis.

III (a).Equation of a Line with Slope

The equation has the form \(y=kx+b\), where \(b=-\dfrac{C}{B}\) and \(k=-\dfrac{A}{B}=\tan\alpha\). Here \(A,\,B,\,C\) are the coefficients of the general equation of the line, and \(\alpha\) is the angle formed by the line with the positive direction of the \(Ox\) axis. The free term \(b\) equals the ordinate of the point where the line meets the \(Oy\) axis.

III (b).Line with Slope \(k\) Through a Point \(M(x_1,y_1)\)

\[y-y_1=k(x-x_1)\]

III (c).Line Through Two Points \(M_1(x_1,y_1)\) and \(M_2(x_2,y_2)\)

\[\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}\]

The slope is \(k=\dfrac{y_2-y_1}{x_2-x_1}\). If \(x_1=x_2\) the line is vertical, \(x=x_1\); if \(y_1=y_2\) it is horizontal, \(y=y_1\).

IV.Equation of a Line in Intercept Form

An equation of the form

\[\frac{x}{a}+\frac{y}{b}=1,\]

where \(a=-\dfrac{C}{A}\) and \(b=-\dfrac{C}{B}\) (with \(A,\,B,\,C\) the coefficients of the general equation of the line), is called the equation of a line in intercept form. Here \(a\) is the abscissa of the point where the line meets the \(Ox\) axis, and \(b\) is the ordinate of the point where it meets the \(Oy\) axis.

V.Normal Equation of a Line

An equation of the form

\[x\cos\varphi+y\sin\varphi-p=0\]

is called the normal equation of a line. Here \(p\) is the length of the perpendicular dropped from the origin onto the line, and \(\varphi\) is the angle formed by this perpendicular with the positive direction of the \(Ox\) axis.

To obtain the normal equation from the general equation \(Ax+By+C=0\), both sides are multiplied by the normalising factor

\[\mu=\pm\frac{1}{\sqrt{A^2+B^2}}.\]

The sign before the radical is chosen so that the condition \(\mu C<0\) holds.

VI (a).Angle Between Lines — Slope Form

The acute angle \(\varphi\) between the lines \(y=k_1x+b_1\) and \(y=k_2x+b_2\) is found by the formula

\[\tan\varphi=\left|\frac{k_2-k_1}{1+k_2k_1}\right|.\]

The lines are parallel when \(k_2=k_1\), and perpendicular when \(k_1=-\dfrac{1}{k_2}\).

VI (b).Angle Between Lines — General Form

If the lines are given by the equations \(A_1x+B_1y+C_1=0\) and \(A_2x+B_2y+C_2=0\), then the angle \(\varphi\) between them is

\[\tan\varphi=\frac{A_1B_2-A_2B_1}{A_1A_2+B_1B_2}.\]

The lines are parallel when \(A_1B_2-A_2B_1=0\) (equivalently \(A_1/A_2=B_1/B_2\)), and perpendicular when \(A_1A_2+B_1B_2=0\).

VII.Intersection of Lines

The coordinates of the intersection point of \(A_1x+B_1y+C_1=0\) and \(A_2x+B_2y+C_2=0\) are found by solving the system

\[\begin{cases}A_1x+B_1y+C_1=0,\\ A_2x+B_2y+C_2=0,\end{cases}\]

which gives

\[x=\frac{\begin{vmatrix}-C_1&B_1\\-C_2&B_2\end{vmatrix}}{\begin{vmatrix}A_1&B_1\\A_2&B_2\end{vmatrix}},\qquad y=\frac{\begin{vmatrix}A_1&-C_1\\A_2&-C_2\end{vmatrix}}{\begin{vmatrix}A_1&B_1\\A_2&B_2\end{vmatrix}}.\]

Comparing the ratios of coefficients:

  • \(\dfrac{A_1}{A_2}\neq\dfrac{B_1}{B_2}\): the lines intersect at one point.
  • \(\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}\neq\dfrac{C_1}{C_2}\): the lines are parallel.
  • \(\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}=\dfrac{C_1}{C_2}\): the lines coincide.

VIII (a).Deviation / Distance of a Point from a Line

The deviation of a point \(M(x_0,y_0)\) from the line \(Ax+By+C=0\) is found by the formula

\[\delta=\frac{Ax_0+By_0+C}{\pm\sqrt{A^2+B^2}},\]

where the sign before the radical is chosen as in the normal equation (§V). The distance \(d\) from the point \(M\) to the line then equals \(d=|\delta|\).

VIII (b).Bisectors of the Angles Between Lines

\[\frac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}}\pm\frac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}}=0\]

The choice of sign is made according to the figure: a point on a bisector is equidistant from both lines, \(|d_1|=|d_2|\).

Two lines I and II with their two angle bisectors; points M₁–M₄ show the signed deviations d₁, d₂ with |d₁|=|d₂|.
Bisectors of the angles between lines I and II (\(|d_1|=|d_2|\)).

IX.Pencil of Lines

If two intersecting lines are given by the equations \(A_1x+B_1y+C_1=0\) and \(A_2x+B_2y+C_2=0\), then the equation

\[A_1x+B_1y+C_1+\lambda(A_2x+B_2y+C_2)=0\]

defines a line passing through the point of intersection of the given lines. Here \(\lambda\) is a numerical parameter. The set of all values of \(\lambda\) defines the pencil of lines, whose centre is the point of intersection of the given lines.

Analytic Geometry in Space

Basic Definitions

The core formulas for planes and straight lines in space — the tools used throughout the problems below.

Plane

I (a).General Equation of a Plane

Every equation of the first degree with respect to the coordinates of a point in space, \(Ax+By+Cz+D=0\), defines a plane; conversely, every plane can be described by an equation of the first degree. Here \(A,\,B,\,C\) are the coordinates of the normal vector:

\[\overline{n}=A\overline{i}+B\overline{j}+C\overline{k}.\]

I (b).Condition for a Point to Lie on a Plane

A point \(M_0(x_0,y_0,z_0)\) lies on the plane \(A_1x+B_1y+C_1z+D_1=0\) precisely when its coordinates satisfy the equation:

\[A_1x_0+B_1y_0+C_1z_0+D_1=0.\]

II.Normal Equation of a Plane

The normal equation of a plane has the form

\[x\cos\alpha+y\cos\beta+z\cos\gamma-p=0,\]

where \(\cos\alpha,\,\cos\beta,\,\cos\gamma\) are the direction cosines of the normal and \(p\) is the distance from the origin to the plane:

\[\cos\alpha=\pm\frac{A}{\sqrt{A^2+B^2+C^2}},\qquad \cos\beta=\pm\frac{B}{\sqrt{A^2+B^2+C^2}},\]
\[\cos\gamma=\pm\frac{C}{\sqrt{A^2+B^2+C^2}},\qquad p=\pm\frac{D}{\sqrt{A^2+B^2+C^2}}.\]

This equation is obtained from the general equation \(Ax+By+Cz+D=0\) by multiplying both sides by the normalising factor \(\mu=\pm\dfrac{1}{\sqrt{A^2+B^2+C^2}}\). The sign of the normalising factor is taken opposite to the sign of the free term \(D\) of the general equation.

III (a).Equation of a Plane in Intercept Form

The intercept form of the equation of a plane is

\[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1,\]

where \(a=-\dfrac{D}{A}\), \(b=-\dfrac{D}{B}\), \(c=-\dfrac{D}{C}\). The numbers \(a,\,b,\,c\) are respectively the abscissa, ordinate, and applicate (the \(x\)-, \(y\)- and \(z\)-intercepts) of the points where the plane meets the coordinate axes.

III (b).Plane Through a Point, Perpendicular to a Vector

The equation of the plane passing through the point \(M_0(x_0,y_0,z_0)\) and perpendicular to the vector \(\overline{n}=A\overline{i}+B\overline{j}+C\overline{k}\) has the form

\[A(x-x_0)+B(y-y_0)+C(z-z_0)=0,\]

or, in vector form, \(\bigl(\overline{n},\,\overline{M_0M}\bigr)=0\).

Plane α with normal vector n through point M₀ and a generic point M(x,y,z)

III (c).Plane Through Three Points

The equation of the plane passing through three given points \(M_1(x_1,y_1,z_1)\), \(M_2(x_2,y_2,z_2)\), \(M_3(x_3,y_3,z_3)\) has the form

\[\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix}=0,\]

or, in vector form, as the vanishing of the scalar triple product \(\bigl(\overline{M_1M_2},\;\overline{M_1M_3},\;\overline{M_1M}\bigr)=0\).

Plane α through three points M₁, M₂, M₃ and a generic point M(x,y,z)

III (d).Pencil of Planes

For an arbitrary value of the parameter \(\lambda\), the equation

\[A_1x+B_1y+C_1z+D_1+\lambda\,(A_2x+B_2y+C_2z+D_2)=0\]

defines a plane passing through the line of intersection of the planes

\[A_1x+B_1y+C_1z+D_1=0 \quad\text{and}\quad A_2x+B_2y+C_2z+D_2=0.\]

III (e).Condition for a Point to Lie on a Plane of the Pencil

The condition for a point \(M_0(x_0,y_0,z_0)\) to lie on a plane belonging to the pencil has the form

\[A_1x_0+B_1y_0+C_1z_0+D_1+\lambda\,(A_2x_0+B_2y_0+C_2z_0+D_2)=0.\]

From this we find the corresponding value \(\lambda_{M_0}\):

\[\lambda_{M_0}=-\frac{A_1x_0+B_1y_0+C_1z_0+D_1}{A_2x_0+B_2y_0+C_2z_0+D_2}.\]

III (f).Plane Through a Point and a Line

The equation of the plane passing through the point \(M_0(x_0,y_0,z_0)\) and the line defined as the intersection of the two planes \(A_1x+B_1y+C_1z+D_1=0\) and \(A_2x+B_2y+C_2z+D_2=0\) has the form

\[A_1x+B_1y+C_1z+D_1+\lambda_{M_0}\,(A_2x+B_2y+C_2z+D_2)=0.\]

IV.Angle Between Two Planes

The angle \(\varphi\) between the planes \(A_1x+B_1y+C_1z+D_1=0\) and \(A_2x+B_2y+C_2z+D_2=0\) is determined by the formula

\[\cos\varphi=\frac{A_1A_2+B_1B_2+C_1C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\,\sqrt{A_2^2+B_2^2+C_2^2}}.\]

The planes are parallel when

\[\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2},\]

perpendicular when

\[A_1A_2+B_1B_2+C_1C_2=0,\]

and intersecting when

\[\frac{A_1}{A_2}\neq\frac{B_1}{B_2}\neq\frac{C_1}{C_2}.\]
Two intersecting planes α and β with the dihedral angle φ between them

V.Deviation of a Point from a Plane

The deviation \(\delta\) of a point \(M_0(x_0,y_0,z_0)\) from the plane \(Ax+By+Cz+D=0\) is found by the formula

\[\delta=\frac{Ax_0+By_0+Cz_0+D}{\pm\sqrt{A^2+B^2+C^2}},\]

where the sign before the radical is taken opposite to the sign of the free term \(D\). The distance from the point \(M_0\) to the plane equals \(d=|\delta|\).

Plane α, an external point M₀, and the perpendicular deviation δ to the plane
Straight Line in Space

VI (a).Line Through Two Points

The equation of the straight line passing through two points \(M_1(x_1,y_1,z_1)\) and \(M_2(x_2,y_2,z_2)\) has the form

\[\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1},\]

or, in vector form,

\[\overline{M_1M}=\lambda\,\overline{M_1M_2},\]

where \(\overline{M_1M}=(x-x_1,\,y-y_1,\,z-z_1)\) and \(\overline{M_1M_2}=(x_2-x_1,\,y_2-y_1,\,z_2-z_1)\).

Straight line L through points M₁, M₂ and a generic point M(x,y,z)

VI (b).Canonical Equation of a Line

The equation of the straight line passing through the point \(M_1(x_1,y_1,z_1)\) parallel to the vector \(\overline{S}=l\overline{i}+m\overline{j}+n\overline{k}\) has the form

\[\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}.\]

This is the canonical equation of the line; the vector \(\overline{S}\) is called the direction vector of the line \(L\).

Straight line L through point M₁ parallel to the direction vector S

VI (c).Parametric Equation of a Line

The parametric equation of a straight line is

\[\begin{cases} x=lt+x_1,\\ y=mt+y_1,\\ z=nt+z_1.\end{cases}\]

It is obtained from the canonical equation by introducing the parameter \(t\): setting \(\dfrac{x-x_1}{l}=\dfrac{y-y_1}{m}=\dfrac{z-z_1}{n}=t\) and solving each ratio for the corresponding coordinate.

VI (d).Line as the Intersection of Two Planes

A straight line in space can be defined by the equations of two planes:

\[\begin{cases} A_1x+B_1y+C_1z+D_1=0,\\ A_2x+B_2y+C_2z+D_2=0,\end{cases}\qquad \overline{n}_1=(A_1,B_1,C_1),\ \ \overline{n}_2=(A_2,B_2,C_2).\]

This is the general equation of a straight line. To bring it to canonical form:

1) Find a point \(M\) on the line by fixing one of the coordinates (say \(z=z_0\)) and solving the system for the remaining two:

\[x_0=\frac{\begin{vmatrix} -C_1z_0-D_1 & B_1 \\ -C_2z_0-D_2 & B_2 \end{vmatrix}}{\begin{vmatrix} A_1 & B_1 \\ A_2 & B_2 \end{vmatrix}},\qquad y_0=\frac{\begin{vmatrix} A_1 & -C_1z_0-D_1 \\ A_2 & -C_2z_0-D_2 \end{vmatrix}}{\begin{vmatrix} A_1 & B_1 \\ A_2 & B_2 \end{vmatrix}}.\]

2) Find the direction vector \(\overline{S}\), parallel to the line (\(\overline{S}\perp\overline{n}_1\) and \(\overline{S}\perp\overline{n}_2\)), as the cross product

\[\overline{S}=\overline{n}_1\times\overline{n}_2=\begin{vmatrix} \overline{i} & \overline{j} & \overline{k} \\ A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \end{vmatrix}=(B_1C_2-B_2C_1)\overline{i}+(A_2C_1-A_1C_2)\overline{j}+(A_1B_2-A_2B_1)\overline{k}=l\overline{i}+m\overline{j}+n\overline{k}.\]

Hence \(\dfrac{x-x_0}{l}=\dfrac{y-y_0}{m}=\dfrac{z-z_0}{n}\) — the canonical form of the line.

VI (e).From Canonical Form to Two Planes

If a line is given by the canonical equation \(\dfrac{x-x_0}{l}=\dfrac{y-y_0}{m}=\dfrac{z-z_0}{n}\), then the pair of equations

\[\begin{cases} \dfrac{x-x_0}{l}=\dfrac{y-y_0}{m},\\[6pt] \dfrac{y-y_0}{m}=\dfrac{z-z_0}{n}\end{cases}\]

defines the same line as the intersection of two planes.

VI (f).Angle Between Two Lines

The angle between two lines in space, given by their canonical equations \(\dfrac{x-x_1}{l_1}=\dfrac{y-y_1}{m_1}=\dfrac{z-z_1}{n_1}\) and \(\dfrac{x-x_2}{l_2}=\dfrac{y-y_2}{m_2}=\dfrac{z-z_2}{n_2}\), is determined by the formula

\[\cos\varphi=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{l_1^2+m_1^2+n_1^2}\,\sqrt{l_2^2+m_2^2+n_2^2}},\]

where \(\overline{S}_1=(l_1,m_1,n_1)\) and \(\overline{S}_2=(l_2,m_2,n_2)\). The lines are parallel when

\[\frac{l_1}{l_2}=\frac{m_1}{m_2}=\frac{n_1}{n_2},\]

and perpendicular when

\[l_1l_2+m_1m_2+n_1n_2=0.\]
Two lines I and II crossing at angle φ, with direction vectors S₁ and S₂

VI (g).Condition for Two Lines to be Coplanar

Two lines given by their canonical equations \(\dfrac{x-x_1}{l_1}=\dfrac{y-y_1}{m_1}=\dfrac{z-z_1}{n_1}\) and \(\dfrac{x-x_2}{l_2}=\dfrac{y-y_2}{m_2}=\dfrac{z-z_2}{n_2}\) are coplanar if and only if

\[\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix}=0.\]

If the numbers \(l_1,m_1,n_1\) are not proportional to \(l_2,m_2,n_2\), this coplanarity condition is the necessary and sufficient condition for the two lines to intersect.

VII.Angle Between a Line and a Plane

The angle between a line \(\dfrac{x-x_1}{l}=\dfrac{y-y_1}{m}=\dfrac{z-z_1}{n}\) and a plane \(Ax+By+Cz+D=0\) is determined by the formula

\[\sin\varphi=\frac{Al+Bm+Cn}{\sqrt{l^2+m^2+n^2}\,\sqrt{A^2+B^2+C^2}},\]

where \(\overline{S}=\{l,m,n\}\) and \(\overline{n}=\{A,B,C\}\). The line is parallel to the plane when

\[Al+Bm+Cn=0,\]

and perpendicular to the plane when

\[\frac{A}{l}=\frac{B}{m}=\frac{C}{n}.\]
Line L piercing plane α at angle φ, showing the normal n and direction vector S

VIII.Point of Intersection of a Line and a Plane

To find the point of intersection of a line \(\dfrac{x-x_0}{l}=\dfrac{y-y_0}{m}=\dfrac{z-z_0}{n}\) with a plane \(Ax+By+Cz+D=0\), write the line in parametric form

\[x=x_0+lt,\qquad y=y_0+mt,\qquad z=z_0+nt,\]

and substitute into the equation of the plane. From it we determine the parameter \(t\):

\[t=-\frac{Ax_0+By_0+Cz_0+D}{Al+Bm+Cn}.\]

Substituting this value back into the parametric equations gives the coordinates \(x,\,y,\,z\) of the intersection point.

  • If \(Al+Bm+Cn\neq0\), the line intersects the plane.
  • If \(Al+Bm+Cn=0\) and \(Ax_0+By_0+Cz_0+D\neq0\), the line is parallel to the plane.
  • If \(Al+Bm+Cn=0\) and \(Ax_0+By_0+Cz_0+D=0\), the line lies in the plane.
Vector Algebra

Basic Definitions

The core operations on vectors given by coordinates — sums and length, the scalar, vector and mixed products, and projection onto an axis — the tools used throughout the problems below.

I. Linear Operations on Vectors

I.1.Coordinates of a Vector

If a vector \(\overline{a}\) has Cartesian rectangular coordinates \(x,\,y,\,z\), then

\[\overline{a}=x\overline{i}+y\overline{j}+z\overline{k}=\{x,\,y,\,z\}.\]

If a vector \(\overline{AB}\) is given by its initial point \(A(x_1,y_1,z_1)\) and terminal point \(B(x_2,y_2,z_2)\), then the coordinates of this vector \(\overline{AB}=\{x,y,z\}\) equal the differences of the like coordinates of the end and the start:

\[x=x_2-x_1;\qquad y=y_2-y_1;\qquad z=z_2-z_1.\]

The zero vector is a vector whose start and end coincide (it has an arbitrary direction).

I.2.Sum and Difference of Vectors

Sum. The sum of two vectors is built either by the triangle rule (place them head to tail) or by the parallelogram rule (draw them from a common origin).

Triangle rule: vector a followed by vector b, resultant c = a + b
Parallelogram rule: vectors a and b from a common origin, diagonal c = a + b

Difference. The difference \(\overline{c}=\overline{a}-\overline{b}\) is the other diagonal of the parallelogram built on \(\overline{a}\) and \(\overline{b}\).

Difference of vectors: parallelogram on a and b, with c = a − b

If \(\overline{a}=\{x_1,y_1,z_1\}\) and \(\overline{b}=\{x_2,y_2,z_2\}\), then

\[\overline{a}\pm\overline{b}=\{x_1\pm x_2;\ y_1\pm y_2;\ z_1\pm z_2\}.\]

If \(\overline{a}=\{x_1,y_1,z_1\}\) and \(\lambda=\text{const}\), then

\[\lambda\overline{a}=\{\lambda x_1,\ \lambda y_1,\ \lambda z_1\}.\]

I.3.Length of a Vector

If \(\overline{a}=\{x,y,z\}\), the length of the vector in an orthonormal basis is

\[|\overline{a}|=\sqrt{x^2+y^2+z^2}.\]

Since

\[\begin{aligned}\cos\alpha&=\cos\bigl(\widehat{\overline{a},\overline{i}}\bigr)=\frac{x}{|\overline{a}|};\\[6pt] \cos\beta&=\cos\bigl(\widehat{\overline{a},\overline{j}}\bigr)=\frac{y}{|\overline{a}|};\\[6pt] \cos\gamma&=\cos\bigl(\widehat{\overline{a},\overline{k}}\bigr)=\frac{z}{|\overline{a}|},\end{aligned}\]

we have \(\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\). The direction cosines of a vector are the coordinates of its ort (unit vector):

\[\overline{a}^{\,0}=\frac{\overline{a}}{|\overline{a}|}=(\cos\alpha,\ \cos\beta,\ \cos\gamma).\]
Scalar Product

I.4.Scalar Product of Vectors

Definition. The scalar product of two vectors is the number equal to the product of their lengths and the cosine of the angle between them:

\[(\overline{a}\cdot\overline{b})=|\overline{a}|\cdot|\overline{b}|\cos\bigl(\widehat{\overline{a}\,\overline{b}}\bigr).\]

Properties of the scalar product:

  1. \(\overline{a}\cdot\overline{b}=\overline{b}\cdot\overline{a}\) — commutativity;
  2. \((\lambda\overline{a})\cdot\overline{b}=\lambda(\overline{a}\,\overline{b})\) — associativity;
  3. \(\overline{a}(\overline{b}+\overline{c})=\overline{a}\,\overline{b}+\overline{a}\,\overline{c}\) — distributivity;
  4. \(\overline{a}\cdot\overline{b}=0\Leftrightarrow\overline{a}\perp\overline{b}\), or \(\overline{a}=0\), or \(\overline{b}=0\);
  5. \(\overline{a}\cdot\overline{a}=|\overline{a}|^2\Rightarrow|\overline{a}|=\sqrt{\overline{a}\cdot\overline{a}}\).

If \(\overline{a}=\{x_1,y_1,z_1\}\) and \(\overline{b}=\{x_2,y_2,z_2\}\), then

\[(\overline{a}\cdot\overline{b})=x_1x_2+y_1y_2+z_1z_2.\]

I.4 (a).Length of \(\overline{a}=\alpha\overline{m}+\beta\overline{n}\)

If \(\overline{a}=\alpha\overline{m}+\beta\overline{n}\), where \(|\overline{m}|,\,|\overline{n}|\) and the angle \(\widehat{\overline{m},\overline{n}}\) between the vectors are known and \(\alpha,\beta=\text{const}\), then

\[\begin{aligned}|\overline{a}|&=\sqrt{\overline{a}\cdot\overline{a}}=\sqrt{(\alpha\overline{m}+\beta\overline{n})(\alpha\overline{m}+\beta\overline{n})}=\\[6pt] &=\sqrt{\alpha^2(\overline{m}\cdot\overline{m})+2\alpha\beta(\overline{m}\cdot\overline{n})+\beta^2(\overline{n}\cdot\overline{n})}=\\[6pt] &=\sqrt{\alpha^2|\overline{m}|^2+2\alpha\beta|\overline{m}||\overline{n}|\cos\bigl(\widehat{\overline{m},\overline{n}}\bigr)+\beta^2|\overline{n}|^2}.\end{aligned}\]

I.4 (b).Scalar Product of Two Combinations

If \(\overline{a}=\alpha\overline{m}+\beta\overline{n}\) and \(\overline{b}=l\overline{m}+d\overline{n}\), where \(|\overline{m}|,\,|\overline{n}|\) and the angle \(\widehat{\overline{m},\overline{n}}\) are known and \(\alpha,\beta,l,d=\text{const}\), then

\[\begin{aligned}(\overline{a}\cdot\overline{b})&=(\alpha\overline{m}+\beta\overline{n})(l\overline{m}+d\overline{n})=\alpha l\,|\overline{m}|^2+\\[6pt] &\quad+(\alpha d+\beta l)\,|\overline{m}||\overline{n}|\cos\bigl(\widehat{\overline{m},\overline{n}}\bigr)+\beta d\,|\overline{n}|^2.\end{aligned}\]

I.4 (c).Angle Between Two Vectors

The angle \(\varphi\) between vectors \(\overline{a}\) and \(\overline{b}\) is computed by the formula

\[\cos\varphi=\frac{(\overline{a}\cdot\overline{b})}{|\overline{a}|\cdot|\overline{b}|}.\]
Two vectors a and b with the angle φ between them

I.4 (d).Angle Between Two Vectors in Coordinates

If \(\overline{a}=\{x_1,y_1,z_1\}\) and \(\overline{b}=\{x_2,y_2,z_2\}\), then

\[\cos\varphi=\frac{(\overline{a}\cdot\overline{b})}{|\overline{a}||\overline{b}|}=\frac{x_1x_2+y_1y_2+z_1z_2}{\sqrt{x_1^2+y_1^2+z_1^2}\,\sqrt{x_2^2+y_2^2+z_2^2}}.\]
Vector Product

I.5.Vector Product of Vectors

Definition. The vector product of a vector \(\overline{a}\) by a vector \(\overline{b}\) is the vector \(\overline{c}\) satisfying the following conditions:

  1. \(|\overline{c}|=|\overline{a}|\cdot|\overline{b}|\sin\bigl(\widehat{\overline{a},\overline{b}}\bigr)\);
  2. \(\overline{c}\perp\overline{a},\ \overline{c}\perp\overline{b}\);
  3. the vector \(\overline{c}\) is directed so that, seen from its end, the shortest rotation from \(\overline{a}\) to \(\overline{b}\) is counterclockwise.

The vector product is denoted \(\overline{a}\times\overline{b}\) or \([\overline{a}\cdot\overline{b}]\).

Properties of the vector product:

  1. \(\overline{a}\times\overline{b}=-\,\overline{b}\times\overline{a}\) — anticommutative;
  2. \(\overline{a}\times\overline{b}=0\) if \(\overline{a}=0\) or \(\overline{b}=0\), or \(\overline{a}\parallel\overline{b}\);
  3. \((\lambda\overline{a})\times\overline{b}=\overline{a}\times(\lambda\overline{b})=\lambda(\overline{a}\times\overline{b})\) — the scalar factor may be taken outside;
  4. \(\overline{a}\times(\overline{b}+\overline{c})=\overline{a}\times\overline{b}+\overline{a}\times\overline{c}\) — distributive over addition.

If \(\overline{a}=\{x_1,y_1,z_1\}\) and \(\overline{b}=\{x_2,y_2,z_2\}\), then

\[\begin{aligned}\overline{a}\times\overline{b}&=\begin{vmatrix}\overline{i}&\overline{j}&\overline{k}\\ x_1&y_1&z_1\\ x_2&y_2&z_2\end{vmatrix}=\\[6pt] &=\overline{i}(y_1z_2-y_2z_1)-\overline{j}(x_1z_2-x_2z_1)+\overline{k}(x_1y_2-x_2y_1).\end{aligned}\]

I.5 (a).Area of a Triangle

The area of the triangle built on the vectors \(\overline{a}=\{x_1,y_1,z_1\}\) and \(\overline{b}=\{x_2,y_2,z_2\}\) is computed by the formula

\[\begin{aligned}S_\Delta&=\tfrac12\,\bigl|[\overline{a},\overline{b}]\bigr|=\tfrac12\left|\begin{vmatrix}\overline{i}&\overline{j}&\overline{k}\\ x_1&y_1&z_1\\ x_2&y_2&z_2\end{vmatrix}\right|=\\[6pt] &=\tfrac12\bigl|(y_1z_2-y_2z_1)\overline{i}+(x_2z_1-x_1z_2)\overline{j}+(x_1y_2-x_2y_1)\overline{k}\bigr|=\\[6pt] &=\tfrac12\sqrt{(y_1z_2-y_2z_1)^2+(x_2z_1-x_1z_2)^2+(x_1y_2-x_2y_1)^2}.\end{aligned}\]

I.5 (b).Vector Product of Two Combinations · Area of a Parallelogram

If \(\overline{a}=\alpha\overline{m}+\beta\overline{n}\) and \(\overline{b}=l\overline{m}+d\overline{n}\), where \(\alpha,\beta,l,d=\text{const}\), then

\[\begin{aligned}[\overline{a},\overline{b}]&=[\alpha\overline{m}+\beta\overline{n},\ l\overline{m}+d\overline{n}]=\\[6pt] &=\alpha l[\overline{m},\overline{m}]+\alpha d[\overline{m},\overline{n}]+\beta l[\overline{n},\overline{m}]+\beta d[\overline{n},\overline{n}]=(\beta l-\alpha d)[\overline{n},\overline{m}],\end{aligned}\]

since \([\overline{m},\overline{n}]=-[\overline{n},\overline{m}]\); \([\overline{m},\overline{m}]=0\); \([\overline{n},\overline{n}]=0\). The area of the parallelogram is

\[\begin{aligned}S_{\text{par}}&=\bigl|[\overline{a},\overline{b}]\bigr|=\bigl|(\beta l-\alpha d)[\overline{n},\overline{m}]\bigr|=\\[6pt] &=|\beta l-\alpha d|\cdot|\overline{n}|\cdot|\overline{m}|\sin\bigl(\widehat{\overline{n},\overline{m}}\bigr).\end{aligned}\]

I.5 (c).A Vector Perpendicular to Two Vectors

If \(\overline{a}=\{x_1,y_1,z_1\}\), \(\overline{b}=\{x_2,y_2,z_2\}\) and \(\overline{c}\perp\overline{a},\ \overline{c}\perp\overline{b}\), then

\[\begin{aligned}\overline{c}=[\overline{a},\overline{b}]&=\begin{vmatrix}\overline{i}&\overline{j}&\overline{k}\\ x_1&y_1&z_1\\ x_2&y_2&z_2\end{vmatrix}=\\[6pt] &=(y_1z_2-y_2z_1)\overline{i}+(x_2z_1-z_2x_1)\overline{j}+(x_1y_2-x_2y_1)\overline{k},\end{aligned}\]

or

\[\overline{c}=X\overline{i}+Y\overline{j}+Z\overline{k},\]

where

\[X=y_1z_2-y_2z_1;\qquad Y=x_2z_1-z_2x_1;\qquad Z=x_1y_2-x_2y_1.\]
Mixed Product

I.6.Mixed Product of Three Vectors

Definition. The mixed product of vectors \(\overline{a},\overline{b}\) and \(\overline{c}\) is the scalar product of the vector \(\overline{a}\times\overline{b}\) with the vector \(\overline{c}\), i.e. \((\overline{a}\times\overline{b})\cdot\overline{c}\).

Properties of the mixed product:

  1. The mixed product equals zero if: (a) at least one of the factors is zero; (b) two of the factors are collinear (lie on parallel lines); (c) the three nonzero vectors are parallel to one and the same plane (coplanarity).
  2. The mixed product does not change if the signs of the vector \((\times)\) and scalar \((\cdot)\) products are interchanged, i.e. \((\overline{a}\times\overline{b})\cdot\overline{c}=\overline{a}\cdot(\overline{b}\times\overline{c})\); it is therefore written simply \(\overline{a}\,\overline{b}\,\overline{c}\).
  3. It does not change under a cyclic permutation of the vectors: \(\overline{a}\,\overline{b}\,\overline{c}=\overline{b}\,\overline{c}\,\overline{a}=\overline{c}\,\overline{a}\,\overline{b}\).
  4. Swapping any two vectors changes only its sign: \(\overline{b}\,\overline{a}\,\overline{c}=-\overline{a}\,\overline{b}\,\overline{c};\ \ \overline{c}\,\overline{b}\,\overline{a}=-\overline{a}\,\overline{b}\,\overline{c};\ \ \overline{a}\,\overline{c}\,\overline{b}=-\overline{a}\,\overline{b}\,\overline{c}.\)

If \(\overline{a}=\{x_1,y_1,z_1\}\); \(\overline{b}=\{x_2,y_2,z_2\}\); \(\overline{c}=\{x_3,y_3,z_3\}\), then

\[\overline{a}\,\overline{b}\,\overline{c}=\begin{vmatrix}x_1&y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3\end{vmatrix}.\]

From these properties it follows that:

(a) the necessary and sufficient condition for the coplanarity of three vectors is \(\overline{a}\,\overline{b}\,\overline{c}=0\), i.e.

\[\begin{vmatrix}x_1&y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3\end{vmatrix}=0;\]

(b) the volume \(V_1\) of the parallelepiped built on \(\overline{a},\overline{b},\overline{c}\) and the volume \(V_2\) of the triangular pyramid (tetrahedron) they form are found by the formulas \(V_1=\bigl|\overline{a}\,\overline{b}\,\overline{c}\bigr|\), or

\[V_1=\text{abs.val}\begin{vmatrix}x_1&y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3\end{vmatrix},\qquad V_2=\tfrac16 V_1=\tfrac16\bigl|\overline{a}\,\overline{b}\,\overline{c}\bigr|,\]

or

\[V_2=\tfrac16\,\text{abs.val}\begin{vmatrix}x_1&y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3\end{vmatrix}.\]
Projection onto an Axis

I.7.Projection of a Vector \(\overline{AB}\) onto an Axis \(l\)

The projection of a vector \(\overline{AB}\) onto an axis \(l\) is the value of the directed segment \(A'B'\) enclosed between the projections of the start and the end of \(\overline{AB}\), taken with a positive sign when \(\overline{A'B'}\) has the direction of the ort of the axis \(l\), and with a negative sign when \(\overline{A'B'}\) and the ort have opposite directions:

\[\text{pr}_l\,\overline{AB}=|\overline{AB}|\cos\bigl(\widehat{\overline{AB},\,l}\bigr)=\frac{(\overline{AB}\cdot\overline{l})}{|\overline{l}|};\]
\[\overline{A'B'}=\frac{\overline{l}}{|\overline{l}|}\cdot\text{pr}_l\,\overline{AB}=\overline{l}\,\frac{(\overline{AB}\cdot\overline{l})}{|\overline{l}|^2}.\]
Projection of vector AB onto axis l, between the feet A′ and B′ of the perpendiculars
Curves of the Second Order

Basic Definitions

The definitions, canonical equations, parameters and construction rules for the ellipse, hyperbola and parabola — the tools used throughout the problems below.

Necessary Information

On the topic “Curves of the Second Order” you should know:

  • the definitions of the ellipse, the hyperbola and the parabola;
  • the canonical equations of these curves;
  • the geometric meaning of the parameters entering the equations, and the relations between the parameters,

and be able to:

  • determine the type of a curve from its equation;
  • derive the canonical equations of second-order curves;
  • find the coordinates of the foci and the other parameters of the curves, and construct the curve.
Ellipse

I.Definition and Canonical Equation

Definition. An ellipse is the set of points of the plane, the sum of whose distances to two given points — called the foci \(F_1\) and \(F_2\) — is a constant (denoted by \(2a\)). Moreover this constant is greater than the distance between the foci.

If the coordinate axes are placed relative to the ellipse as shown in Fig. 1, and the foci lie on the \(Ox\) axis at equal distances from the origin at the points \(F_1(c;0)\), \(F_2(-c;0)\), one obtains the simplest (canonical) equation of the ellipse:

\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\]

where \(a\) is the semi-major and \(b\) the semi-minor axis of the ellipse, and \(a,\,b\) and \(c\) (\(c\) is half the distance between the foci) are related by \(a^2=b^2+c^2\).

The shape of the ellipse (its measure of “flattening”) is characterised by its eccentricity: \(\varepsilon=\dfrac{c}{a}\) (since \(c<a\), we have \(\varepsilon<1\)).

The lines \(D_1:\ x=-\dfrac{a}{\varepsilon}\) and \(D_2:\ x=+\dfrac{a}{\varepsilon}\), perpendicular to the major axis and passing at distance \(\dfrac{a}{\varepsilon}\) from the centre, are called the directrices of the ellipse.

Ellipse with foci F₁, F₂, a point M(x,y) with focal radii r₁, r₂, and the directrices D₁, D₂.

I.1.Location of the Ellipse and Its Parameters

\[\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1;\qquad O'(x_0,y_0)\ \text{— centre.}\]

1) \(a>b,\ c=\sqrt{a^2-b^2},\ F_1(x_0+c;\,y_0),\ F_2(x_0-c;\,y_0)\);

2) \(a<b,\ c=\sqrt{b^2-a^2},\ F_1(x_0;\,y_0+c),\ F_2(x_0;\,y_0-c)\).

Ellipse a>b, centre O′(x₀,y₀), semi-axes a, b, distance c to the foci F₁, F₂, shifted axes X′, Y′.

I.2.Eccentricity

\[\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1.\]

1) \(a>b;\ \varepsilon=\dfrac{c}{a}=\dfrac{\sqrt{a^2-b^2}}{a}\) — eccentricity;

2) \(a<b;\ \varepsilon=\dfrac{c}{b}=\dfrac{\sqrt{b^2-a^2}}{b}\) — eccentricity.

Ellipse b>a, foci on the vertical axis, horizontal directrices D₁, D₂.

I.3.Equations of the Directrices

1) \(a>b\): \(\quad D_2:\ x=x_0+\dfrac{a}{\varepsilon};\qquad D_1:\ x=x_0-\dfrac{a}{\varepsilon}\);

2) \(b>a\): \(\quad D_2:\ y=y_0+\dfrac{b}{\varepsilon};\qquad D_1:\ y=y_0-\dfrac{b}{\varepsilon}\).

Hyperbola

II.Definition and Canonical Equation

Definition. A hyperbola is the set of points of the plane, the absolute value of the difference of whose distances to two given points — called the foci — is a constant (denoted by \(2a\)); moreover this constant is less than the distance between the foci.

If the foci of the hyperbola are placed at the points \(F_1(c;0)\) and \(F_2(-c;0)\), one obtains the canonical equation of the hyperbola

\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\qquad\text{where}\quad b^2=c^2-a^2.\]

The points \(A_1(a;0)\) and \(A_2(-a;0)\) are called the vertices of the hyperbola. The segment \(A_1A_2\) with \(|A_1A_2|=2a\) is called the real (transverse) axis of the hyperbola, and the segment \(B_1B_2\) with \(|B_1B_2|=2b\) the imaginary axis.

The hyperbola has two asymptotes, whose equations are

\[y=\pm\frac{b}{a}\,x.\]

The ratio \(\varepsilon=\dfrac{c}{a}>1\) is called the eccentricity of the hyperbola.

The equation \(\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1\) is also an equation of a hyperbola, but the real axis of this hyperbola is the segment of the \(Oy\) axis of length \(2b\).

The two hyperbolas \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\) and \(\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1\) have the same semi-axes and the same asymptotes, but the real axis of one serves as the imaginary axis of the other and vice versa. Such two hyperbolas are called conjugate.

The lines \(D_1:\ x=-\dfrac{a}{\varepsilon}\) and \(D_2:\ x=+\dfrac{a}{\varepsilon}\), perpendicular to the real axis and passing at distance \(\dfrac{a}{\varepsilon}\) from the centre, are called the directrices of the hyperbola.

Hyperbola with vertices A₁, A₂, imaginary-axis points B₁, B₂, foci F₁, F₂, focal radii r₁, r₂, asymptotes and directrices D₁, D₂.

II.1.Location of the Hyperbola

1) \(\dfrac{(x-x_0)^2}{a^2}-\dfrac{(y-y_0)^2}{b^2}=1;\ O'(x_0;y_0)\) — centre, \(c=\sqrt{a^2+b^2}\), \(F_1(x_0+c;\,y_0)\), \(F_2(x_0-c;\,y_0)\) (Fig. 5);

Horizontal hyperbola, centre O′(x₀,y₀), semi-axes a, b, foci F₁, F₂, asymptotes.

2) \(\dfrac{(y-y_0)^2}{b^2}-\dfrac{(x-x_0)^2}{a^2}=1;\ O'(x_0;\,y_0)\) — centre, \(c=\sqrt{a^2+b^2}\), \(F_1(x_0;\,y_0-c)\), \(F_2(x_0;\,y_0+c)\) (Fig. 6).

Vertical hyperbola, centre O′, foci F₁, F₂ on the vertical axis, asymptotes.

II.2.Eccentricity

1) \(\dfrac{(x-x_0)^2}{a^2}-\dfrac{(y-y_0)^2}{b^2}=1;\ \varepsilon=\dfrac{c}{a}=\dfrac{\sqrt{a^2+b^2}}{a}\) — eccentricity.

2) \(\dfrac{(y-y_0)^2}{b^2}-\dfrac{(x-x_0)^2}{a^2}=1;\ \varepsilon=\dfrac{c}{b}=\dfrac{\sqrt{a^2+b^2}}{b}\) — eccentricity.

II.3.Equations of the Directrices of the Hyperbola

1) \(\dfrac{(x-x_0)^2}{a^2}-\dfrac{(y-y_0)^2}{b^2}=1,\)

\[D_2:\ x=x_0+\dfrac{a}{\varepsilon};\qquad D_1:\ x=x_0-\dfrac{a}{\varepsilon};\]

2) \(\dfrac{(y-y_0)^2}{b^2}-\dfrac{(x-x_0)^2}{a^2}=1,\)

\[D_2:\ y=y_0+\dfrac{b}{\varepsilon};\qquad D_1:\ y=y_0-\dfrac{b}{\varepsilon}.\]

II.4.Equation of the Asymptotes

\[y-y_0=\pm\frac{b\,(x-x_0)}{a}\qquad\text{— equation of the asymptotes of the hyperbola.}\]
Parabola

III.Definition and Canonical Equation

Definition. A parabola is the set of points of the plane equidistant from a given point — called the focus — and a given line — called the directrix.

If the directrix of the parabola is the line \(D:\ x=-\dfrac{p}{2}\), and the focus is the point \(F\left(\dfrac{p}{2};\,0\right)\), then the equation of the parabola has the form

\[y^2=2px,\qquad\text{where}\quad p>0.\]
Parabola y²=2px with focus F, a point M, focal radius r and the directrix D.

III.1.Location of the Parabola and Its Parameters

1) \((x-x_0)^2=2p(y-y_0);\ O'(x_0,y_0)\) — vertex, \(p>0,\ F(x_0,\,y_0+\tfrac{p}{2})\) (Fig. 8) or \(p<0,\ F(x_0,\,y_0+\tfrac{p}{2})\) (Fig. 9).

Upward parabola (p>0), vertex O′, focus F at distance p/2.
Downward parabola (p<0), vertex O′, focus F.

2) \((y-y_0)^2=2p(x-x_0);\ O'(x_0,y_0)\) — vertex, \(p>0,\ F(x_0+\tfrac{p}{2},\,y_0)\) (Fig. 10) or \(p<0,\ F(x_0+\tfrac{p}{2},\,y_0)\) (Fig. 11).

Rightward parabola (p>0), vertex O′, focus F.
Leftward parabola (p<0), vertex O′, focus F.

III.2.Equation of the Directrix of the Parabola

1) \((x-x_0)^2=2p(y-y_0),\)

\[p>0\ \ D:\ y=y_0-\dfrac{p}{2},\qquad p<0\ \ D:\ y=y_0-\dfrac{p}{2};\]

2) \((y-y_0)^2=2p(x-x_0),\)

\[p>0\ \ D:\ x=x_0-\dfrac{p}{2},\qquad p<0\ \ D:\ x=x_0-\dfrac{p}{2}.\]
General Equation of a Second-Order Curve

IV.1.Type from the Product \(A\cdot C\)

For the investigation of second-order curves, whose general equation has the form \(Ax^2+Cy^2+2Dx+2Ey+F=0\), one considers the product \(A\cdot C\):

  • if \(A\cdot C>0\), we have an ellipse;
  • if \(A\cdot C<0\), we have a hyperbola;
  • if \(A\cdot C=0\), we have a parabola.

IV.2.Completing the Square

In the investigation of second-order curves whose equation is written in general form, the “procedure of completing the square” is useful. Completing the square of the equation \(Ax^2+Cy^2+2Dx+2Ey+F=0\), we obtain

\[A\!\left(x^2+2\frac{D}{A}x\right)+C\!\left(y^2+2\frac{E}{C}y\right)+F=0\]

or

\[A\!\left(x+\frac{D}{A}\right)^2+C\!\left(y+\frac{E}{C}\right)^2=\frac{D^2}{A}+\frac{E^2}{C}-F.\]

Denote: \(x_0=-\dfrac{D}{A},\ y_0=-\dfrac{E}{C}\),

\[a^2=\frac{\dfrac{D^2}{A}+\dfrac{E^2}{C}-F}{A},\qquad b^2=\frac{\dfrac{D^2}{A}+\dfrac{E^2}{C}-F}{C}.\]

IV.2 (a).Elliptic Type \((A\cdot C>0)\)

If \(A\cdot C>0\), the equation defines a curve of elliptic type. Moreover:

a) if \(\dfrac{D^2}{A}+\dfrac{E^2}{C}-F<0\), we have an imaginary ellipse;

b) if \(\dfrac{D^2}{A}+\dfrac{E^2}{C}-F=0\), we have the point \(\left(-\dfrac{D}{A};\,-\dfrac{E}{C}\right)\);

c) if \(\dfrac{D^2}{A}+\dfrac{E^2}{C}-F>0\), then

\[\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1\]

and we have the canonical form of an ellipse.

IV.2 (b).Hyperbolic Type \((A\cdot C<0)\)

If \(A\cdot C<0\), the equation defines a curve of hyperbolic type. Moreover:

a) if \(\dfrac{D^2}{A}+\dfrac{E^2}{C}-F>0\) or \(\dfrac{D^2}{A}+\dfrac{E^2}{C}-F<0\), then

\[\frac{(x-x_0)^2}{a^2}-\frac{(y-y_0)^2}{b^2}=1\quad\text{or}\quad\frac{(y-y_0)^2}{b^2}-\frac{(x-x_0)^2}{a^2}=1,\]

i.e. the canonical form of a hyperbola;

b) if \(\dfrac{D^2}{A}+\dfrac{E^2}{C}-F=0\), then, taking the signs of \(A\) and \(C\) into account, we have \(y-y_0=\pm\dfrac{b}{a}(x-x_0)\), i.e. a pair of intersecting lines.

IV.2 (c).Parabolic Type

a) If \(C=0\), the general equation \(Ax^2+Cy^2+2Dx+2Ey+F=0\) defines a curve of parabolic type. Completing the square, we have:

\[A\!\left(x+\frac{D}{A}\right)^2=-2E\!\left(y+\frac{\dfrac{D^2}{A}-F}{-2E}\right).\]

Denote \(x_0=-\dfrac{D}{A},\ y_0=-\dfrac{F-\dfrac{D^2}{A}}{2E},\ p=-\dfrac{E}{A}\). Then \((x-x_0)^2=2p(y-y_0)\), i.e. the canonical form of a parabola;

b) if \(A=0\), then \(Ax^2+Cy^2+2Dx+2Ey+F=0\), i.e. a curve of parabolic type. Completing the square, we have:

\[C\!\left(y+\frac{E}{C}\right)^2=-2D\!\left(x+\frac{F-\dfrac{E^2}{C}}{2D}\right).\]

Denote \(x_0=-\dfrac{F-\dfrac{E^2}{C}}{2D},\ y_0=-\dfrac{E}{C},\ p=-\dfrac{D}{C}\). Then \((y-y_0)^2=2p(x-x_0)\), i.e. the canonical form of a parabola.

Incomplete Equations of Curves

V.1.Equations of the Form \(y=b+\sqrt{ax^2+cx+d}\)

Consider an equation of the form \(y=b+\sqrt{ax^2+cx+d}\). This equation is equivalent to the system

\[\begin{cases} y-b\ge 0\\ (y-b)^2=ax^2+cx+d\end{cases}\sim\begin{cases} y\ge b\\ (y-b)^2-a\!\left(x+\dfrac{c}{2a}\right)^2=d-\dfrac{c^2}{4a}.\end{cases}\]

V.1 (a). If \(d-\dfrac{c^2}{4a}>0\) and \(a>0\), the equation \((y-b)^2-a\left(x+\tfrac{c}{2a}\right)^2=d-\tfrac{c^2}{4a}\) defines the part of a hyperbola lying in the half-plane \(y\ge b\) (Fig. 12). We construct the part of the hyperbola above the line \(y=b\).

Part of a hyperbola (upper branch) lying above the line y=b.

V.1 (b). If \(d-\dfrac{c^2}{4a}>0\) and \(a<0\), the equation defines the part of an ellipse lying in the half-plane \(y\ge b\) (Fig. 13). We construct the part of the ellipse above the line \(y=b\).

Upper half of an ellipse lying above the line y=b.

V.1 (c). If \(d-\dfrac{c^2}{4a}<0\) and \(a>0\), the equation defines the part of a hyperbola lying in the half-plane \(y\ge b\) (Fig. 14).

Two upper arcs of a hyperbola lying above the line y=b.

V.1 (d). If \(d-\dfrac{c^2}{4a}<0\) and \(a<0\), the equation defines an imaginary ellipse.

V.1 (e). If \(a=0\), the equation \(y=b+\sqrt{cx+d}\) is equivalent to the system \(\begin{cases}y-b\ge 0\\ (y-b)^2=c\left(x+\tfrac{d}{c}\right)\end{cases}\), which defines the part of a parabola lying in the half-plane \(y\ge b\) (Figs. 15, 16).

Part of a parabola (c>0) above the line y=b.
Part of a parabola (c<0) above the line y=b.

V.1 (f). If \(d-\dfrac{c^2}{4a}=0\) and \(a>0\), the equation defines a pair of intersecting lines \(y-b=\pm\sqrt{a}\left(x+\tfrac{c}{2a}\right)\), which lie in the half-plane \(y\ge b\).

V.1 (g). If \(d-\dfrac{c^2}{4a}=0\) and \(a<0\), the equation defines the point \(\left(-\tfrac{c}{2a};\,b\right)\), provided it lies in the half-plane \(y\ge b\).

V.2.Equations of the Form \(y=b-\sqrt{ax^2+cx+d}\)

Consider an equation of the form \(y=b-\sqrt{ax^2+cx+d}\). This equation is equivalent to the system

\[\begin{cases} y-b\le 0\\ (y-b)^2-a\!\left(x+\dfrac{c}{2a}\right)^2=d-\dfrac{c^2}{4a},\end{cases}\]

since only positive values of the root are considered.

V.2 (a). If \(d-\dfrac{c^2}{4a}>0\) and \(a>0\), the equation defines the part of a hyperbola lying in the half-plane \(y\le b\) (Fig. 17).

Part of a hyperbola (lower branch) lying below the line y=b.

V.2 (b). If \(d-\dfrac{c^2}{4a}>0\) and \(a<0\), the equation defines the part of an ellipse lying in the half-plane \(y\le b\) (Fig. 18).

Lower half of an ellipse lying below the line y=b.

V.2 (c). If \(d-\dfrac{c^2}{4a}<0\) and \(a>0\), the equation defines the part of a hyperbola lying in the half-plane \(y\le b\) (Fig. 19).

Two lower arcs of a hyperbola lying below the line y=b.

V.2 (d). If \(d-\dfrac{c^2}{4a}<0\) and \(a<0\), the equation defines an imaginary ellipse.

V.2 (e). If \(a=0\), the equation \(y=b-\sqrt{cx+d}\) is equivalent to the system \(\begin{cases}y\le b\\ (y-b)^2=c\left(x+\tfrac{d}{c}\right)\end{cases}\), which defines the part of a parabola lying in the half-plane \(y\le b\) (Figs. 20, 21).

Part of a parabola (c>0) below the line y=b.
Part of a parabola (c<0) below the line y=b.

V.2 (f). If \(d-\dfrac{c^2}{4a}=0\) and \(a>0\), the equation defines a pair of intersecting lines \(y-b=\pm\sqrt{a}\left(x+\tfrac{c}{2a}\right)\), which lie in the half-plane \(y\le b\).

V.2 (g). If \(d-\dfrac{c^2}{4a}=0\) and \(a<0\), the equation defines the point \(\left(-\tfrac{c}{2a};\,b\right)\), provided it lies in the half-plane \(y\le b\).

Analytic Geometry in the Plane · Problem 1

General Equation of a Line

Problem Statement

Find the general equation of line BC, given B(−2, −1) and C(4, 3).

1
/1

Problem Solved!

Complete Solution
Analytic Geometry in the Plane · Problem 5

Lines & Equal Distances

Equation of a line · distance from a point to a line · angle between two lines — with one subtle case the obvious method misses.

Problem Statement

Calculate the angle between lines a and b, both passing through A(1, 3) and equally distant from B(−2, −1) and C(4, 3).

Choose the algorithm to solve the problem
🤔

Before you compute — think

Algorithm A writes the unknown line as y−3 = k(x−1) and sets the distance from B equal to the distance from C. The whole equation is built around the slope k.

Hint: what kind of line through A has no slope k at all? Could the distance equation ever produce it?

Algorithm B · Geometric Symmetry

Why Symmetry Works

Before we solve step by step, here is the geometric insight behind Algorithm B.

Key insight: two cases of equidistance

A line through A is equidistant from B and C in exactly two geometric cases:
Case 1
Parallel to BC
A line parallel to BC is the same perpendicular distance from every point on BC — so it is equally far from both B and C.
→ gives line a
Case 2
Through midpoint of BC
The midpoint E of segment BC is by definition equidistant from B and C. Any line through both A and E satisfies the condition.
→ gives line b
7-step solution plan
1
Find the equation of line BC
2
Read off the slope of BC
3
Write line a — parallel to BC through A
4
Find midpoint E of segment BC
5
Write line b — through E and A
6
Write the equations of lines a and b
7
Calculate the angle between a and b
2 / 8
8
/8

Problem Solved!

Algorithm B — Complete Solution
Analytic Geometry in the Plane · Problem 2

The Perpendicular from a Point to a Line

Equation of a line · slope of a line · the perpendicularity condition for slopes.

Problem Statement

On the plane three points are given: A(1, 3), B(−2, −1), C(4, 3). Write the general equation of the perpendicular AD, dropped from point A onto line BC.

Choose the algorithm to solve the problem
🤔

Before you compute — think

Algorithm B step 1 says "find a vector parallel to AD" — but that is a goal, not an instruction. It never says how.

Hint: a valid algorithm is a list of explicit, elementary steps a person (or computer) can follow blindly. Which sub-steps does "find a parallel vector" quietly leave out?

Algorithm A · Through the Slope of BC

Perpendicular Means Negative-Reciprocal Slopes

Before solving step by step, here is the idea Algorithm A is built on.

Key insight: the perpendicularity condition

Two lines are perpendicular exactly when their slopes are negative reciprocals:
Condition
kAD = −1 / kBC
Since ADBC, once you know the slope of BC the slope of AD is fixed — flip it and change its sign.
→ gives the slope of AD
Then
Point A pins it down
AD passes through the given point A(1, 3). A slope plus a point determines the line uniquely.
→ gives the equation of AD
4-step solution plan
1
Write the equation of line BC
2
Find the slope of BC
3
Find the slope of AD (perpendicular to BC)
4
Write the equation of AD through A
2 / 5
5
/5

Problem Solved!

Algorithm A — Complete Solution
Analytic Geometry in the Plane · Problem 3

Bisectors of the Angles Between Two Lines

Equation of a line · perpendicular slopes · the angle-bisector formula.

Problem Statement

On the plane three points are given: A(1, 3), B(−2, −1), C(4, 3). Write the equations of the bisectors of the angles formed by lines BC and AD. The lines are perpendicular.

Choose the algorithm to solve the problem
🤔

Before you compute — think

Algorithm B step 2 says "find a vector perpendicular to BC" — but that is a goal, not an instruction. It never says how.

Hint: a valid algorithm lists explicit, elementary steps a person (or computer) can follow blindly. What method does "find a perpendicular vector" quietly assume you already know?

Algorithm A · Through the Slope of BC

From the Perpendicular to its Bisectors

Before solving step by step, here are the two ideas Algorithm A rests on.

Key insight

First reconstruct the perpendicular AD, then bisect the right angle it makes with BC:
Step 1–4
kAD = −1 / kBC
Since ADBC, the slope of AD is the negative reciprocal of the slope of BC; through A this fixes line AD.
→ gives line AD
Step 5
Equal distances
A point on a bisector is equidistant from both lines, giving the formula with the two signs — one bisector per sign.
→ gives both bisectors
5-step solution plan
1
Write the equation of line BC
2
Find the slope of BC
3
Find the slope of AD (perpendicular to BC)
4
Write the equation of AD through A
5
Write the equations of the two bisectors
2 / 6
6
/6

Problem Solved!

Algorithm A — Complete Solution
Analytic Geometry in the Plane · Problem 4

Lines Equidistant from Two Points

Equation of a line · parallel lines · midpoint of a segment — find both lines through a point that are equally far from two others.

Problem Statement

On the plane three points are given: A(1, 3), B(−2, −1), C(4, 3). Write the equations of lines a and b, passing through point A and equally distant from points B and C.

Choose the algorithm to solve the problem
🤔

Before you compute — think

Algorithm B jumps from the midpoint E straight to line a. But line a must be parallel to BC — and writing it needs the slope of BC.

Hint: which step of Algorithm B ever finds the direction (slope) of BC? If none does, can line a be written at all?

Algorithm A · Through Line BC

Two Ways to Be Equidistant

Before solving step by step, here is the geometric idea behind Algorithm A.

Key insight: two cases of equidistance

A line through A is equidistant from B and C in exactly two geometric cases:
Case 1
Parallel to BC
A line parallel to BC is the same perpendicular distance from every point on BC — so it is equally far from both B and C.
→ gives line a
Case 2
Through midpoint of BC
The midpoint E of segment BC is by definition equidistant from B and C. Any line through both A and E satisfies the condition.
→ gives line b
6-step solution plan
1
Write the equation of line BC
2
Find the slope of BC
3
Write line a — parallel to BC through A
4
Find midpoint E of segment BC
5
Write line b — through E and A
6
Write the equations of lines a and b
2 / 7
7
/7

Problem Solved!

Algorithm A — Complete Solution
Section

Analytic Geometry in the Plane

Focused on two-dimensional space, this section covers the geometric properties of lines in a coordinate plane. It explains how to find the coordinates of a point that divides a line segment in a given ratio, explores the various forms of linear equations, and demonstrates how to calculate the shortest distance from a point to a line.

Section

Analytic Geometry in Space

Extends the methods of plane geometry into three dimensions. Covers the equations of planes and straight lines in space, distances from a point to a plane, and angles between geometric objects.

Section

Vector Algebra

Works with vectors given by their coordinates in space. Covers the linear operations, the length and direction of a vector, and the three products — scalar, vector and mixed — used to find angles, areas, volumes and normals.

Analytic Geometry in Space · Problem 1

The Line of Intersection of Two Planes

Arrangement of two planes · direction vector of a line · canonical equations of a line in space.

Problem Statement

Having verified that the planes α: 2x + 4y − 4z + 12 = 0 and β: x − 3y + z = 0 intersect, compose the canonical equations of their line of intersection (line a).

Choose the algorithm to solve the problem
🤔

Before you compute — think

Algorithm B jumps straight to a direction vector — but it never checks that the two planes actually meet in a line.

Hint: if the planes were parallel or coincident there would be no line a at all. The problem itself says "having verified that the planes intersect". Which step does Algorithm B quietly skip?

Algorithm A · Establish, Then Build

A Line in Space Is a Point Plus a Direction

Before solving step by step, here is the idea Algorithm A is built on.

Key insight: a line is fixed by one point and one direction

The intersection line lies inside both planes, so its geometry is pinned down by the two normals:
Direction
S = nα × nβ
Line a lies in both planes, so it is perpendicular to both normal vectors. Their cross product gives a vector along the line.
→ gives the direction of a
Anchor
One shared point M1
Any point satisfying both plane equations lies on a. A single common solution anchors the line in space.
→ gives a point on a
4-step solution plan
1
Investigate the arrangement of the planes
2
Find a point M1 on line a
3
Find the direction vector S of line a
4
Write the canonical equations of line a
2 / 5
5
/5

Problem Solved!

Algorithm A — Complete Solution
Analytic Geometry in Space · Problem 2

The Perpendicular from a Point to a Plane

Normal of a plane · direction of a perpendicular · canonical equations of a line in space.

Problem Statement

Write the canonical equations of the perpendicular AB dropped from the point A(1, 1, 0) onto the plane α: 2x + 4y − 4z + 12 = 0.

Choose the algorithm to solve the problem
🤔

Before you compute — think

Algorithm B's first step is to find where AB meets α — but AB is precisely the line you're trying to construct.

Hint: you can't intersect a line with a plane before you have the line. Is finding the foot an instruction you can follow, or a result that already assumes the answer?

Algorithm A · Straight Down the Normal

A Line Is a Point Plus a Direction

Before solving step by step, here is the idea Algorithm A is built on.

Key insight: the perpendicular is already pinned down

You don't need the foot of the perpendicular — a direction and a point are handed to you:
Direction
S = nα
A line perpendicular to a plane runs along that plane's normal. Read the normal straight off α from the coefficients of x, y, z.
→ gives the direction of AB
Anchor
The given point A
The perpendicular starts at A(1, 1, 0), which is given. One point and one direction fix a line completely.
→ gives a point on AB
2-step solution plan
1
Find the direction vector S of AB
2
Write the canonical equations of AB
2 / 5
5
/5

Problem Solved!

Algorithm A — Complete Solution
Analytic Geometry in Space · Problem 3

A Plane Perpendicular to Two Planes

Normals of planes · cross product · equation of a plane through a point.

Problem Statement

Write the equation of the plane γ passing through the point A(1, 1, 0) and perpendicular to the two planes α: 2x + 4y − 4z + 12 = 0 and β: x − 3y + z = 0.

Choose the algorithm to solve the problem
🤔

Before you compute — think

Algorithm B's step 2 says "find nγ from the perpendicularity conditions" — but that names a goal, not a procedure you can carry out.

Hint: written as conditions, nγ ⟂ nα and nγ ⟂ nβ are two equations in three unknowns. What single elementary operation actually produces a vector perpendicular to two given ones?

Algorithm A · Normals, Then Cross Product

The Normal of γ Is Perpendicular to Both

Before solving step by step, here is the idea Algorithm A is built on.

Key insight: the cross product builds the normal of γ

If γ is perpendicular to both planes, its normal is perpendicular to both of their normals:
Normal
nγ = nα × nβ
γ ⟂ α and γ ⟂ β means nγ is perpendicular to both nα and nβ. The cross product returns exactly such a vector.
→ gives the normal of γ
Anchor
The given point A
The plane must pass through A(1, 1, 0). A point and a normal fix a plane completely.
→ gives a point on γ
3-step solution plan
1
Find the normals of α and β
2
Find the normal nγ = nα × nβ
3
Write the equation of plane γ
2 / 5
5
/5

Problem Solved!

Algorithm A — Complete Solution
Analytic Geometry in Space · Problem 4

A Plane Through a Point and a Line

Pencil of planes · fitting the parameter λ · equation of a plane.

Problem Statement

Write the equation of the plane δ passing through the point A(1, 1, 0) and the line given as the intersection of the planes α: 2x + 4y − 4z + 12 = 0 and β: x − 3y + z = 0.

Choose the algorithm to solve the problem
🤔

Before you compute — think

Algorithm B does reach the answer — but each of its lines is a whole procedure in disguise, not a single step.

Hint: "find the direction and a point of the line" is the entire Problem 1 (solve a system + a cross product); "find AM1 and the normal" is another cross product. Is there a route that skips building the line at all?

Algorithm A · The Pencil of Planes

One Equation Sweeps Every Plane Through the Line

Before solving step by step, here is the idea Algorithm A is built on.

Key insight: the pencil turns about the line α ∩ β

You never need to build the line — the pencil already contains every plane through it:
The pencil
α + λβ = 0
For each value of λ, this equation is a plane containing the line α ∩ β. As λ varies, it sweeps through all of them.
→ all planes through the line
Fit to A
One value of λ
Requiring the plane to pass through A(1, 1, 0) is one linear equation in λ. Its solution picks out plane δ.
→ selects plane δ
3-step solution plan
1
Write the pencil of planes
2
Find λ so the plane passes through A
3
Write the equation of plane δ
2 / 5
5
/5

Problem Solved!

Algorithm A — Complete Solution
Analytic Geometry in Space · Problem 5

The Distance from a Point to a Plane

Point-in-plane test · the distance formula in space.

Problem Statement

Compute the distance d from the point A(1, 1, 0) to the plane 2x + 4y − 4z + 12 = 0.

Choose the algorithm to solve the problem
🤔

Before you compute — think

Algorithm B does give the right distance — but it takes the long way, building the whole perpendicular and its foot just to measure one segment.

Hint: each of its three lines is a sub-problem on its own (a line, an intersection, a length). Is there a formula that returns the distance from the point and the plane directly?

Algorithm A · Straight from the Formula

One Formula Gives the Distance

Before solving step by step, here is the idea Algorithm A is built on.

Key insight: the distance is a single formula

No construction is needed — substitute the point into the plane and normalise:
Formula
d = |Ax₀+By₀+Cz₀+D| / √(A²+B²+C²)
The numerator is the plane's left side evaluated at the point; the denominator is the length of its normal.
→ gives the distance directly
Sanity check
Is A on the plane?
If substituting A gives 0, the point lies on the plane and the distance is 0. Here it gives 18, so the distance is positive.
→ confirms d ≠ 0
2-step solution plan
1
Check whether A lies on the plane
2
Compute the distance from the formula
2 / 5
5
/5

Problem Solved!

Algorithm A — Complete Solution
Vector Algebra · Problem 1

The Angle Between an Edge and a Diagonal

Coordinates of vectors · lengths · scalar product · angle between vectors.

Problem Statement

The parallelepiped ABCDA′B′C′D′ is built on the vectors AB(6, 2, 2), AD(1, 1, −1), AA′(−2, 3, 1). Compute the angle φ between edge A′D′ and diagonal DB′.

Choose the algorithm to solve the problem
🤔

Algorithm B is invalid

Its last step reads sin φ from the scalar product — but by the definition in I.4 the scalar product equals |a||b| cos φ, so it can only give the cosine.

That step has no basis in the theory, so Algorithm B cannot reach the angle. (The sine would require the vector product, I.5.)

Algorithm A · Through the Scalar Product

The Scalar Product Gives the Cosine

Before solving step by step, here is the idea Algorithm A is built on.

Key insight

The whole solution rests on these two facts:
Vectors
A′D′ and DB′
Both are found by subtracting the coordinates of the endpoints.
→ two vectors in coordinates
Angle
cos φ = (a·b)/(|a||b|)
The scalar product carries the cosine of the angle between the vectors.
→ gives cos φ directly
4-step solution plan
1
Coordinates of A′D′ and DB′
2
Lengths |A′D′| and |DB′|
3
Scalar product (A′D′·DB′)
4
cos φ
2 / 5
5
/5

Problem Solved!

Algorithm A — Complete Solution
Vector Algebra · Problem 2

The Area of a Triangular Section

Coordinates of vectors · vector product · area of a triangle.

Problem Statement

The parallelepiped ABCDA′B′C′D′ is built on AB(6, 2, 2), AD(1, 1, −1), AA′(−2, 3, 1). Compute the area S of the section EB′C, where E is the midpoint of edge D′C′.

Choose the algorithm to solve the problem
🤔

Algorithm B works — A is preferred here

By I.5 (a) the two sides may be taken from any common vertex of the triangle. Algorithm A uses C (CB′, CE); Algorithm B uses B′ (B′E, B′C). Their vector products have equal magnitude, so both give the same area.

So B is not wrong — same steps, same area. Algorithm A is preferred only because it works from vertex C, whose vectors CB′, CE (and CC′ = AA′) are reused in Problems 3 and 4. Since B reaches the answer with the same effort, it is marked amber (a hint), not an error.

Algorithm A · From Vertex C

Half a Vector Product Is a Triangle’s Area

Before solving step by step, here is the idea Algorithm A is built on.

Key insight

The whole solution rests on these two facts:
Sides
CB′ and CE
Two edges of the triangle sharing vertex C, found from coordinates.
→ two sides of the triangle
Area
S = ½ |CB′ × CE|
The vector product’s length is the parallelogram area; half of it is the triangle.
→ gives the area S
3-step solution plan
1
Coordinates of CB′ and CE
2
Vector product [CB′ × CE]
3
Area S = ½|CB′ × CE|
2 / 4
4
/4

Problem Solved!

Algorithm A — Complete Solution
Vector Algebra · Problem 3

The Volume of a Tetrahedron

Coordinates of vectors · mixed product · volume of a tetrahedron.

Problem Statement

The parallelepiped ABCDA′B′C′D′ is built on AB(6, 2, 2), AD(1, 1, −1), AA′(−2, 3, 1). Compute the volume of the tetrahedron EB′C′C, where E is the midpoint of edge D′C′.

Choose the algorithm to solve the problem
🤔

Algorithm B is valid but inefficient

B builds the base area with a vector product (I.5), then the tetrahedron’s height, then combines them — three detours for what I.6 does in one step.

By I.6 (b) the volume is V = ⅙|CB′ × CE · CC′| directly. Algorithm A uses that.

Algorithm A · Through the Mixed Product

The Mixed Product Is Six Volumes

Before solving step by step, here is the idea Algorithm A is built on.

Key insight

The whole solution rests on these two facts:
Edges
CB′, CE, CC′
Three edges of the tetrahedron meeting at C, from coordinates.
→ three edge vectors
Volume
V = ⅙ |a b c|
The mixed product is the parallelepiped volume; one sixth is the tetrahedron.
→ gives the volume
3-step solution plan
1
Coordinates of CB′, CE, CC′
2
Mixed product CB′ × CE · CC′
3
Volume V = ⅙|mixed product|
2 / 4
4
/4

Problem Solved!

Algorithm A — Complete Solution
Vector Algebra · Problem 4

A Unit Normal to a Plane

Coordinates of vectors · vector product · unit normal vector.

Problem Statement

The parallelepiped ABCDA′B′C′D′ is built on AB(6, 2, 2), AD(1, 1, −1), AA′(−2, 3, 1). Find the coordinates of the unit vector n₀ perpendicular to the plane B′CE, where E is the midpoint of edge D′C′.

Choose the algorithm to solve the problem
🤔

Algorithm A is the step-by-step route

By I.5 (c) the normal can be formed from two sides of the plane drawn from any vertex. Algorithm A uses C (CB′, CE); Algorithm B uses B′ (B′C, B′E). Their products point opposite ways, but I.5 (c) fixes the unit normal only up to sign (±), so both give the same n₀.

The difference is the layout. Algorithm B merges two steps into one — it asks you to find the vector product and its length together. Algorithm A keeps them separate (find [CB′ × CE], then its length, then normalise), so each elementary step can be followed and checked on its own. That is why A is the route to use.

Algorithm A · From Vertex C

The Vector Product Points Across the Plane

Before solving step by step, here is the idea Algorithm A is built on.

Key insight

The whole solution rests on these two facts:
Normal
CB′ × CE
The vector product of two sides is perpendicular to the plane they span.
→ a normal vector
Unit
n₀ = ± N / |N|
Divide the normal by its length; the sign is free (two directions).
→ the unit normal
4-step solution plan
1
Coordinates of CB′ and CE
2
Vector product [CB′ × CE]
3
Length |CB′ × CE|
4
Unit vector n₀
2 / 5
5
/5

Problem Solved!

Algorithm A — Complete Solution
Vector Algebra · Problem 5

The Foot of a Perpendicular

Length · unit vector · scalar product · projection of a vector.

Problem Statement

The parallelepiped ABCDA′B′C′D′ is built on AB(6, 2, 2), AD(1, 1, −1), AA′(−2, 3, 1). Find the coordinates of the vector AF, where F is the foot of the perpendicular dropped from point A′ onto edge AB.

Choose the algorithm to solve the problem
🤔

Algorithm A is the efficient choice

Algorithm B first computes the length |AA′| and the cosine of the angle (I.4 (c)) — introducing square roots that cancel out in the final step.

Algorithm A skips all of that: by I.7 the projection is (AA′·AB)/|AB| straight from the dot product, and AF follows at once — no square roots. That is why A is the efficient route.

Algorithm A · Through the Projection

A Projection Lands the Foot of the Perpendicular

Before solving step by step, here is the idea Algorithm A is built on.

Key insight

The whole solution rests on these two facts:
Direction
AB₀ = AB / |AB|
The unit vector along AB gives the direction of the foot.
→ unit direction of AB
Projection
AF = (AA′·AB₀) AB₀
The signed projection of AA′ onto AB, placed along AB₀, is the vector AF.
→ gives AF
5-step solution plan
1
Length |AB|
2
Unit vector AB₀
3
Scalar product (AA′·AB)
4
Projection of AA′ onto AB
5
Coordinates of AF
2 / 6
6
/6

Problem Solved!

Algorithm A — Complete Solution
Section

Curves of the Second Order

Recognise and analyse the ellipse, hyperbola and parabola from a general second-degree equation. Determine the type from the product A·C, complete the square to reach the canonical form, and find the centre, foci, eccentricity, directrices, asymptotes — and construct the curve.

Curves of the Second Order · Problem 1

An Ellipse in General Position

Type of a second-order curve · completing the square · foci, eccentricity, directrices · construction.

Problem Statement

Investigate the equation 4x² + 9y² − 8x − 36y + 4 = 0 — determine the type of curve it defines and find its parameters.

Choose the algorithm to solve the problem
🤔

Before you commit — think

Algorithm B stops at the sign of A·C: it names the family — ellipse, hyperbola or parabola — and calls the problem done.

Hint: the sign of A·C only classifies the quadratic part. Without completing the square you never locate the centre, so the foci, eccentricity and directrices stay unknown — and you cannot tell a genuine curve from a degenerate one (a single point, an empty “imaginary” locus, or a pair of lines). What does completing the square reveal here?

Algorithm A · Reduce to Canonical Form

Complete the Square, Then Read Off Everything

Before solving step by step, here is the idea Algorithm A is built on.

Key insight: the canonical form settles both the nature and the parameters

The sign of A·C names the family; completing the square then relocates the centre and exposes everything else.
Classify
sign of A·C
The product of the coefficients of and fixes the type: positive → elliptic, negative → hyperbolic, zero → parabolic.
→ the family of the curve
Locate
complete the square
Rewriting in canonical form moves the centre to (x₀,y₀) and gives the semi-axes — and reveals any degeneration into a point, an empty locus or lines.
→ centre, axes & every parameter
six-step solution plan
1
Determine the type of the curve
2
Reduce the equation to canonical form
3
Find the coordinates of the foci
4
Find the eccentricity
5
Write the equations of the directrices
6
Construct the curve
2 / 7
7
/7

Problem Solved!

Algorithm A — Complete Solution
Curves of the Second Order · Problem 2

A Degenerate Ellipse — a Single Point

Type of a second-order curve · completing the square · a degenerate (point) locus.

Problem Statement

Investigate the equation x² + y² − 2x + 4y + 5 = 0 — determine the type of curve it defines and find its parameters.

Choose the algorithm to solve the problem
🤔

Before you commit — think

Algorithm B stops at the sign of A·C: it names the family — ellipse, hyperbola or parabola — and calls the problem done.

Hint: the sign of A·C only classifies the quadratic part. Without completing the square you never locate the centre, so the foci, eccentricity and directrices stay unknown — and you cannot tell a genuine curve from a degenerate one (a single point, an empty “imaginary” locus, or a pair of lines). What does completing the square reveal here?

Algorithm A · Reduce to Canonical Form

Complete the Square, Then Read Off Everything

Before solving step by step, here is the idea Algorithm A is built on.

Key insight: the canonical form settles both the nature and the parameters

The sign of A·C names the family; completing the square then relocates the centre and exposes everything else.
Classify
sign of A·C
The product of the coefficients of and fixes the type: positive → elliptic, negative → hyperbolic, zero → parabolic.
→ the family of the curve
Locate
complete the square
Rewriting in canonical form moves the centre to (x₀,y₀) and gives the semi-axes — and reveals any degeneration into a point, an empty locus or lines.
→ centre, axes & every parameter
two-step solution plan
1
Determine the type of the curve
2
Reduce the equation to canonical form
2 / 3
3
/3

Problem Solved!

Algorithm A — Complete Solution
Curves of the Second Order · Problem 3

A Degenerate Ellipse — an Imaginary Locus

Type of a second-order curve · completing the square · an imaginary (empty) ellipse.

Problem Statement

Investigate the equation x² + y² + 4x + 5 = 0 — determine the type of curve it defines and find its parameters.

Choose the algorithm to solve the problem
🤔

Before you commit — think

Algorithm B stops at the sign of A·C: it names the family — ellipse, hyperbola or parabola — and calls the problem done.

Hint: the sign of A·C only classifies the quadratic part. Without completing the square you never locate the centre, so the foci, eccentricity and directrices stay unknown — and you cannot tell a genuine curve from a degenerate one (a single point, an empty “imaginary” locus, or a pair of lines). What does completing the square reveal here?

Algorithm A · Reduce to Canonical Form

Complete the Square, Then Read Off Everything

Before solving step by step, here is the idea Algorithm A is built on.

Key insight: the canonical form settles both the nature and the parameters

The sign of A·C names the family; completing the square then relocates the centre and exposes everything else.
Classify
sign of A·C
The product of the coefficients of and fixes the type: positive → elliptic, negative → hyperbolic, zero → parabolic.
→ the family of the curve
Locate
complete the square
Rewriting in canonical form moves the centre to (x₀,y₀) and gives the semi-axes — and reveals any degeneration into a point, an empty locus or lines.
→ centre, axes & every parameter
two-step solution plan
1
Determine the type of the curve
2
Reduce the equation to canonical form
2 / 3
3
/3

Problem Solved!

Algorithm A — Complete Solution
Curves of the Second Order · Problem 4

A Hyperbola in General Position

Type of a second-order curve · completing the square · foci, eccentricity, directrices, asymptotes · construction.

Problem Statement

Investigate the equation x² − 9y² + 2x + 36y − 44 = 0 — determine the type of curve it defines and find its parameters.

Choose the algorithm to solve the problem
🤔

Before you commit — think

Algorithm B stops at the sign of A·C: it names the family — ellipse, hyperbola or parabola — and calls the problem done.

Hint: the sign of A·C only classifies the quadratic part. Without completing the square you never locate the centre, so the foci, eccentricity and directrices stay unknown — and you cannot tell a genuine curve from a degenerate one (a single point, an empty “imaginary” locus, or a pair of lines). What does completing the square reveal here?

Algorithm A · Reduce to Canonical Form

Complete the Square, Then Read Off Everything

Before solving step by step, here is the idea Algorithm A is built on.

Key insight: the canonical form settles both the nature and the parameters

The sign of A·C names the family; completing the square then relocates the centre and exposes everything else.
Classify
sign of A·C
The product of the coefficients of and fixes the type: positive → elliptic, negative → hyperbolic, zero → parabolic.
→ the family of the curve
Locate
complete the square
Rewriting in canonical form moves the centre to (x₀,y₀) and gives the semi-axes — and reveals any degeneration into a point, an empty locus or lines.
→ centre, axes & every parameter
seven-step solution plan
1
Determine the type of the curve
2
Reduce the equation to canonical form
3
Find the coordinates of the foci
4
Find the eccentricity
5
Write the equations of the directrices
6
Write the equations of the asymptotes
7
Construct the curve
2 / 8
8
/8

Problem Solved!

Algorithm A — Complete Solution
Curves of the Second Order · Problem 5

A Degenerate Hyperbola — a Pair of Lines

Type of a second-order curve · completing the square · a degenerate (line-pair) locus.

Problem Statement

Investigate the equation 16y² − 9x² − 32y + 18x + 7 = 0 — determine the type of curve it defines and find its parameters.

Choose the algorithm to solve the problem
🤔

Before you commit — think

Algorithm B stops at the sign of A·C: it names the family — ellipse, hyperbola or parabola — and calls the problem done.

Hint: the sign of A·C only classifies the quadratic part. Without completing the square you never locate the centre, so the foci, eccentricity and directrices stay unknown — and you cannot tell a genuine curve from a degenerate one (a single point, an empty “imaginary” locus, or a pair of lines). What does completing the square reveal here?

Algorithm A · Reduce to Canonical Form

Complete the Square, Then Read Off Everything

Before solving step by step, here is the idea Algorithm A is built on.

Key insight: the canonical form settles both the nature and the parameters

The sign of A·C names the family; completing the square then relocates the centre and exposes everything else.
Classify
sign of A·C
The product of the coefficients of and fixes the type: positive → elliptic, negative → hyperbolic, zero → parabolic.
→ the family of the curve
Locate
complete the square
Rewriting in canonical form moves the centre to (x₀,y₀) and gives the semi-axes — and reveals any degeneration into a point, an empty locus or lines.
→ centre, axes & every parameter
two-step solution plan
1
Determine the type of the curve
2
Reduce the equation to canonical form
2 / 3
3
/3

Problem Solved!

Algorithm A — Complete Solution