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The core formulas for straight lines in the coordinate plane — the tools used throughout the problems below.
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:
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:
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.
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, \(y=y_1\).
An equation of the form
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.
An equation of the form
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
The sign before the radical is chosen so that the condition \(\mu C<0\) holds.
The acute angle \(\varphi\) between the lines \(y=k_1x+b_1\) and \(y=k_2x+b_2\) is found by the formula
The lines are parallel when \(k_2=k_1\), and perpendicular when \(k_1=-\dfrac{1}{k_2}\).
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
The lines are parallel when \(A_1B_2-A_2B_1=0\) (equivalently \(A_1/A_2=B_1/B_2\)), and perpendicular when \(A_1A_2+B_1B_2=0\).
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
which gives
Comparing the ratios of coefficients:
The deviation of a point \(M(x_0,y_0)\) from the line \(Ax+By+C=0\) is found by the formula
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 \(d=|\delta|\).
The choice of sign is made according to the figure: a point on a bisector is equidistant from both lines, \(|d_1|=|d_2|\).
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
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.
The core formulas for planes and straight lines in space — the tools used throughout the problems below.
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:
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:
The normal equation of a plane has the form
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:
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.
The intercept form of the equation of a plane is
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.
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
or, in vector form, \(\bigl(\overline{n},\,\overline{M_0M}\bigr)=0\).
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
or, in vector form, as the vanishing of the scalar triple product \(\bigl(\overline{M_1M_2},\;\overline{M_1M_3},\;\overline{M_1M}\bigr)=0\).
For an arbitrary value of the parameter \(\lambda\), the equation
defines a plane passing through the line of intersection of the planes
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
From this we find the corresponding value \(\lambda_{M_0}\):
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
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
The planes are parallel when
perpendicular when
and intersecting when
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
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 \(d=|\delta|\).
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
or, in vector form,
where \(\overline{M_1M}=(x-x_1,\,y-y_1,\,z-z_1)\) and \(\overline{M_1M_2}=(x_2-x_1,\,y_2-y_1,\,z_2-z_1)\).
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
This is the canonical equation of the line; the vector \(\overline{S}\) is called the direction vector of the line \(L\).
The parametric equation of a straight line is
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.
A straight line in space can be defined by the equations of two planes:
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:
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
Hence \(\dfrac{x-x_0}{l}=\dfrac{y-y_0}{m}=\dfrac{z-z_0}{n}\) — the canonical form of the line.
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
defines the same line as the intersection of two planes.
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
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
and perpendicular when
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
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.
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
where \(\overline{S}=\{l,m,n\}\) and \(\overline{n}=\{A,B,C\}\). The line is parallel to the plane when
and perpendicular to the plane when
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
and substitute into the equation of the plane. From it we determine the parameter \(t\):
Substituting this value back into the parametric equations gives the coordinates \(x,\,y,\,z\) of the intersection point.
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.
If a vector \(\overline{a}\) has Cartesian rectangular coordinates \(x,\,y,\,z\), then
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:
The zero vector is a vector whose start and end coincide (it has an arbitrary direction).
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).
Difference. The difference \(\overline{c}=\overline{a}-\overline{b}\) is the other diagonal of the parallelogram built on \(\overline{a}\) and \(\overline{b}\).
If \(\overline{a}=\{x_1,y_1,z_1\}\) and \(\overline{b}=\{x_2,y_2,z_2\}\), then
If \(\overline{a}=\{x_1,y_1,z_1\}\) and \(\lambda=\text{const}\), then
If \(\overline{a}=\{x,y,z\}\), the length of the vector in an orthonormal basis is
Since
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):
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:
Properties of the scalar product:
If \(\overline{a}=\{x_1,y_1,z_1\}\) and \(\overline{b}=\{x_2,y_2,z_2\}\), then
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
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
The angle \(\varphi\) between vectors \(\overline{a}\) and \(\overline{b}\) is computed by the formula
If \(\overline{a}=\{x_1,y_1,z_1\}\) and \(\overline{b}=\{x_2,y_2,z_2\}\), then
Definition. The vector product of a vector \(\overline{a}\) by a vector \(\overline{b}\) is the vector \(\overline{c}\) satisfying the following conditions:
The vector product is denoted \(\overline{a}\times\overline{b}\) or \([\overline{a}\cdot\overline{b}]\).
Properties of the vector product:
If \(\overline{a}=\{x_1,y_1,z_1\}\) and \(\overline{b}=\{x_2,y_2,z_2\}\), then
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
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
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
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
or
where
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:
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
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.
(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
or
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:
The definitions, canonical equations, parameters and construction rules for the ellipse, hyperbola and parabola — the tools used throughout the problems below.
On the topic “Curves of the Second Order” you should know:
and be able to:
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:
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.
1) \(a>b,\ c=\sqrt{a^2-b^2},\ F_1(x_0+c;\,y_0),\ F_2(x_0-c;\,y_0)\);
2) \(a<b,\ c=\sqrt{b^2-a^2},\ F_1(x_0;\,y_0+c),\ F_2(x_0;\,y_0-c)\).
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.
1) \(a>b\): \(\quad D_2:\ x=x_0+\dfrac{a}{\varepsilon};\qquad D_1:\ x=x_0-\dfrac{a}{\varepsilon}\);
2) \(b>a\): \(\quad D_2:\ y=y_0+\dfrac{b}{\varepsilon};\qquad D_1:\ y=y_0-\dfrac{b}{\varepsilon}\).
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
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
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.
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);
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).
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.
1) \(\dfrac{(x-x_0)^2}{a^2}-\dfrac{(y-y_0)^2}{b^2}=1,\)
2) \(\dfrac{(y-y_0)^2}{b^2}-\dfrac{(x-x_0)^2}{a^2}=1,\)
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
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).
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).
1) \((x-x_0)^2=2p(y-y_0),\)
2) \((y-y_0)^2=2p(x-x_0),\)
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 \(A\cdot C\):
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
or
Denote: \(x_0=-\dfrac{D}{A},\ y_0=-\dfrac{E}{C}\),
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
and we have the canonical form of an ellipse.
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
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.
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:
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:
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.
Consider an equation of the form \(y=b+\sqrt{ax^2+cx+d}\). This equation is equivalent to the system
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 \(y=b\).
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 \(y=b\).
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).
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).
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 \(y\ge b\).
Consider an equation of the form \(y=b-\sqrt{ax^2+cx+d}\). This equation is equivalent to the system
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).
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).
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).
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).
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 \(y\le b\).
Find the general equation of line BC, given B(−2, −1) and C(4, 3).
Equation of a line · distance from a point to a line · angle between two lines — with one subtle case the obvious method misses.
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).
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?
Before we solve step by step, here is the geometric insight behind Algorithm B.
Equation of a line · slope of a line · the perpendicularity condition for slopes.
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.
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?
Before solving step by step, here is the idea Algorithm A is built on.
Equation of a line · perpendicular slopes · the angle-bisector formula.
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.
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?
Before solving step by step, here are the two ideas Algorithm A rests on.
Equation of a line · parallel lines · midpoint of a segment — find both lines through a point that are equally far from two others.
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.
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?
Before solving step by step, here is the geometric idea behind Algorithm A.
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.
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.
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.
Arrangement of two planes · direction vector of a line · canonical equations of a line in space.
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).
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?
Before solving step by step, here is the idea Algorithm A is built on.
Normal of a plane · direction of a perpendicular · canonical equations of a line in space.
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.
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?
Before solving step by step, here is the idea Algorithm A is built on.
Normals of planes · cross product · equation of a plane through a point.
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.
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?
Before solving step by step, here is the idea Algorithm A is built on.
Pencil of planes · fitting the parameter λ · equation of a plane.
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.
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?
Before solving step by step, here is the idea Algorithm A is built on.
Point-in-plane test · the distance formula in space.
Compute the distance d from the point A(1, 1, 0) to the plane 2x + 4y − 4z + 12 = 0.
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?
Before solving step by step, here is the idea Algorithm A is built on.
Coordinates of vectors · lengths · scalar product · angle between vectors.
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′.
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.)
Before solving step by step, here is the idea Algorithm A is built on.
Coordinates of vectors · vector product · area of a triangle.
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′.
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.
Before solving step by step, here is the idea Algorithm A is built on.
Coordinates of vectors · mixed product · volume of a tetrahedron.
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′.
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.
Before solving step by step, here is the idea Algorithm A is built on.
Coordinates of vectors · vector product · unit normal vector.
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′.
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.
Before solving step by step, here is the idea Algorithm A is built on.
Length · unit vector · scalar product · projection of a vector.
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.
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.
Before solving step by step, here is the idea Algorithm A is built on.
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.
Type of a second-order curve · completing the square · foci, eccentricity, directrices · construction.
Investigate the equation 4x² + 9y² − 8x − 36y + 4 = 0 — determine the type of curve it defines and find its parameters.
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?
Before solving step by step, here is the idea Algorithm A is built on.
Type of a second-order curve · completing the square · a degenerate (point) locus.
Investigate the equation x² + y² − 2x + 4y + 5 = 0 — determine the type of curve it defines and find its parameters.
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?
Before solving step by step, here is the idea Algorithm A is built on.
Type of a second-order curve · completing the square · an imaginary (empty) ellipse.
Investigate the equation x² + y² + 4x + 5 = 0 — determine the type of curve it defines and find its parameters.
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?
Before solving step by step, here is the idea Algorithm A is built on.
Type of a second-order curve · completing the square · foci, eccentricity, directrices, asymptotes · construction.
Investigate the equation x² − 9y² + 2x + 36y − 44 = 0 — determine the type of curve it defines and find its parameters.
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?
Before solving step by step, here is the idea Algorithm A is built on.
Type of a second-order curve · completing the square · a degenerate (line-pair) locus.
Investigate the equation 16y² − 9x² − 32y + 18x + 7 = 0 — determine the type of curve it defines and find its parameters.
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?
Before solving step by step, here is the idea Algorithm A is built on.