Basic Definitions

A second-order curve is a plane curve whose Cartesian coordinates satisfy an algebraic equation of degree two. Its real, imaginary, nondegenerate, and degenerate cases are classified by reduction to canonical form and by the invariants of the associated quadratic form.

This section covers the ellipse, the hyperbola, and the parabola: their definitions, their canonical equations, and the geometric meaning of each parameter. It shows how to determine a curve's type from its equation, derive its canonical form, and find its foci and other parameters in order to construct it.

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 in the figure, 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 (conjugate) 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, $c=\sqrt{a^2+b^2}$, $F_1(x_0+c;\,y_0)$, $F_2(x_0-c;\,y_0)$;

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

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, $p>0,\ F(x_0,\,y_0+\tfrac{p}{2})$ or $p<0,\ F(x_0,\,y_0+\tfrac{p}{2})$.

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

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

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

For an axis-aligned second-order curve, the equation has the form $Ax^2+Cy^2+2Dx+2Ey+F=0$. Because there is no $xy$ term, the coordinate axes are aligned with the principal directions of the quadratic part. The sign of $AC$ identifies the broad family: $AC>0$ is elliptic type, $AC<0$ is hyperbolic type, and $AC=0$ is parabolic type. This sign alone does not determine whether the curve is degenerate or whether it has real points; completing the square is still required.

In the investigation of second-order curves whose equation is written in general form, the “procedure of completing the square” is useful. This reduction assumes $A\neq0$ and $C\neq0$ (a central conic); the case $AC=0$ is treated in IV.2 (c). Completing the square of the equation $Ax^2+Cy^2+2Dx+2Ey+F=0$, we obtain

or

Denote $x_0=-\dfrac{D}{A},\ y_0=-\dfrac{E}{C}$, and write $R=\dfrac{D^2}{A}+\dfrac{E^2}{C}-F$ for the right-hand side, so that

If $AC>0$, the equation defines a curve of elliptic type. Since $AC>0$, $a^2=\dfrac{R}{A}$ and $b^2=\dfrac{R}{C}$ have the same sign. Then:

a) if $\dfrac{R}{A}<0$ (so $a^2<0$), we have an imaginary ellipse (no real points);

b) if $R=0$, we have the single point $\left(-\dfrac{D}{A};\,-\dfrac{E}{C}\right)$;

c) if $\dfrac{R}{A}>0$, then

the canonical form of an ellipse.

The test is the sign of $R/A$, not of $R$: when $A,C<0$, multiplying the equation by $-1$ first makes $A,C>0$ and reverses the sign of $R$, giving the same answer. For example $-x^2-y^2+1=0$ has $R=-1$ but $R/A=1$ — the real unit circle.

If $AC<0$, the equation defines a curve of hyperbolic type. From completing the square, $A(x-x_0)^2+C(y-y_0)^2=R$. Since $AC<0$, $R/A$ and $R/C$ have opposite signs, so $a^2$ and $b^2$ cannot both be taken positive — work from this form directly:

a) if $R\neq0$, divide by $R$; the two terms then have opposite signs, and writing the positive term first gives

the canonical form of a hyperbola;

b) if $R=0$, then $A(x-x_0)^2+C(y-y_0)^2=0$ with $AC<0$ splits into a pair of intersecting lines $y-y_0=\pm\sqrt{-A/C}\,(x-x_0)$.

If $AC=0$ and exactly one of $A$, $C$ is nonzero, the quadratic part is a single perfect square; if $A=C=0$ there is no quadratic term and the equation is linear (case (c) below). A single square gives a parabola only when the matching linear term is present as well:

a) if $C=0$ with $A\neq0$ and $E\neq0$, then $Ax^2+2Dx+2Ey+F=0$ is a non-degenerate parabola. Completing the square:

Denote $x_0=-\dfrac{D}{A},\ y_0=-\dfrac{F-\dfrac{D^2}{A}}{2E},\ p=-\dfrac{E}{A}$; then $(x-x_0)^2=2p(y-y_0)$, the canonical form of a parabola. If instead $E=0$, the equation reduces to $A\left(x+\dfrac{D}{A}\right)^2=\dfrac{D^2}{A}-F$: two parallel lines, one repeated line, or no real points, according to the sign of $\dfrac{D^2/A-F}{A}$;

b) if $A=0$ with $C\neq0$ and $D\neq0$, symmetrically:

Denote $x_0=-\dfrac{F-\dfrac{E^2}{C}}{2D},\ y_0=-\dfrac{E}{C},\ p=-\dfrac{D}{C}$; then $(y-y_0)^2=2p(x-x_0)$. If instead $D=0$, the equation reduces to $C\left(y+\dfrac{E}{C}\right)^2=\dfrac{E^2}{C}-F$: two parallel lines, one repeated line, or no real points;

c) if $A=C=0$, the equation $2Dx+2Ey+F=0$ is of the first degree — a single line, or (when $D=E=0$) the empty set or the whole plane, not a conic.

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

The completed-square form on the right assumes $a\neq0$; the case $a=0$ is (e) below.

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 $y\ge b$. 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 $y\ge b$. 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 $y\ge b$.

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

V.1 (e). If $a=0$ and $c\neq0$, the equation $y=b+\sqrt{cx+d}$ is equivalent to $\begin{cases}y-b\ge 0\\ (y-b)^2=c\left(x+\tfrac{d}{c}\right)\end{cases}$, which defines the part of a parabola lying in the half-plane $y\ge b$. If $a=c=0$, the equation reads $y=b+\sqrt{d}$: the horizontal line $y=b+\sqrt{d}$ when $d\ge0$, and no real points when $d<0$.

V.1 (f). If $d-\dfrac{c^2}{4a}=0$ and $a>0$, then $(y-b)^2=a\left(x+\tfrac{c}{2a}\right)^2$; together with $y\ge b$ this gives $y-b=\sqrt{a}\,\bigl|x+\tfrac{c}{2a}\bigr|$ — the two rays in the half-plane $y\ge b$ that are the upper halves of the lines $y-b=\pm\sqrt{a}\left(x+\tfrac{c}{2a}\right)$, meeting at $\left(-\tfrac{c}{2a};\,b\right)$.

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 the principal square root is nonnegative. The completed-square form assumes $a\neq0$; the case $a=0$ is (e) below.

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 $y\le b$.

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 $y\le b$.

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 $y\le b$.

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

V.2 (e). If $a=0$ and $c\neq0$, the equation $y=b-\sqrt{cx+d}$ is equivalent to $\begin{cases}y\le b\\ (y-b)^2=c\left(x+\tfrac{d}{c}\right)\end{cases}$, which defines the part of a parabola lying in the half-plane $y\le b$. If $a=c=0$, the equation reads $y=b-\sqrt{d}$: the horizontal line $y=b-\sqrt{d}$ when $d\ge0$, and no real points when $d<0$.

V.2 (f). If $d-\dfrac{c^2}{4a}=0$ and $a>0$, then $(y-b)^2=a\left(x+\tfrac{c}{2a}\right)^2$; together with $y\le b$ this gives $y-b=-\sqrt{a}\,\bigl|x+\tfrac{c}{2a}\bigr|$ — the two rays in the half-plane $y\le b$ that are the lower halves of the lines $y-b=\pm\sqrt{a}\left(x+\tfrac{c}{2a}\right)$, meeting at $\left(-\tfrac{c}{2a};\,b\right)$.

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 $y\le b$.