General Conics, calculus lesson

The equation $5x^2 - 6xy + 5y^2 = 8$ hides an ellipse, but it is tilted, so none of the standard forms fit. Every conic, wherever it sits and however it is turned, is described by a second degree equation. The job here is to read that equation and say what the curve is.

Key terms

General quadratic equation

Definition A general second degree equation in $x$ and $y$ is

Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,

where $A, B, C, D, E, F$ are real constants and $A$ , $B$ , $C$ are not all zero. Its solution set is a conic, possibly degenerate, a pair of parallel lines, or empty.

In plain words Any mix of $x^2$ , $xy$ , $y^2$ , $x$ , $y$ and a constant, with at least one second degree term.

Example The hyperbola $xy = 1$ is the case $B = 1$ , $F = -1$ , with all other coefficients zero. The circle $x^2 + y^2 = 4$ is $A = C = 1$ , $F = -4$ .

Shifted conics

Definition If an equation $G(x, y) = 0$ describes a curve $\mathcal{C}$ , then $G(x - h, y - k) = 0$ describes the curve obtained by moving every point of $\mathcal{C}$ by $h$ horizontally and $k$ vertically. For example

\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1

is the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ moved so its center is $(h, k)$ , with foci, vertices and directrices moved by the same amount.

In plain words Replacing $x$ by $x - h$ slides the picture right by $h$ , not left. Shape, size and eccentricity stay the same.

Example The equation

\frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{4} = 1

describes an ellipse with center $(2, -1)$ , $a = 3$ , $b = 2$ and $c = \sqrt{9 - 4} = \sqrt{5}$ , so its foci are $(2 \pm \sqrt{5}, -1)$ .

Completing the square

Definition For any real $b$ ,

x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4}.

When $B = 0$ , completing the square in $x$ (if $A \ne 0$ ) and in $y$ (if $C \ne 0$ ) puts the equation in a shifted standard form or shows that it is degenerate.

In plain words Group the $x$ terms and the $y$ terms, then turn each group into a perfect square plus a constant.

Example Take $4x^2 + 9y^2 - 16x + 18y - 11 = 0$ . Then

4(x^2 - 4x) + 9(y^2 + 2y) = 11,

4(x - 2)^2 - 16 + 9(y + 1)^2 - 9 = 11,

so $4(x - 2)^2 + 9(y + 1)^2 = 36$ . Dividing by $36$ gives the ellipse

\frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{4} = 1

from the previous section.

Rotation of axes

Definition Rotating the coordinate axes counterclockwise by an angle $\theta$ gives new coordinates $(X, Y)$ for the same point, related to $(x, y)$ by

x = X\cos\theta - Y\sin\theta, \qquad y = X\sin\theta + Y\cos\theta,

and equivalently

X = x\cos\theta + y\sin\theta, \qquad Y = -x\sin\theta + y\cos\theta.

Distances and angles are the same in both systems, so a conic keeps its type and shape.

In plain words The curve stays where it is. You turn the grid so the curve lines up with the new axes.

Example With $\theta = \frac{\pi}{4}$ the formulas become

x = \frac{X - Y}{\sqrt{2}}, \qquad y = \frac{X + Y}{\sqrt{2}}, \qquad xy = \frac{X^2 - Y^2}{2}.

So the curve $xy = 1$ becomes

\frac{X^2}{2} - \frac{Y^2}{2} = 1,

a hyperbola with $a = b = \sqrt{2}$ .

Removing the cross term

Definition Suppose $B \ne 0$ . Rotating the axes by an angle $\theta$ with $0 < \theta < \frac{\pi}{2}$ and

\cot 2\theta = \frac{A - C}{B}

turns the equation into

A'X^2 + C'Y^2 + D'X + E'Y + F = 0,

with no $XY$ term and the same constant $F$ . The term $Bxy$ is the cross term. For a nondegenerate conic that is not a circle, it is nonzero exactly when the axes of the conic are tilted relative to the coordinate axes.

In plain words The $xy$ term means the curve is turned. The formula tells you how far to turn the axes to straighten it.

Example For $5x^2 - 6xy + 5y^2 - 8 = 0$ , $A = C$ gives $\cot 2\theta = 0$ , so $\theta = \frac{\pi}{4}$ . Using $x^2 + y^2 = X^2 + Y^2$ and $xy = \frac{X^2 - Y^2}{2}$ ,

5(X^2 + Y^2) - 3(X^2 - Y^2) = 8, \qquad \frac{X^2}{4} + Y^2 = 1,

an ellipse with semimajor axis $2$ along the line $y = x$ .

Discriminant classification

Definition The discriminant $B^2 - 4AC$ does not change under rotation of axes. Let

\Delta = \det \begin{pmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{pmatrix}.

If $\Delta \ne 0$ , the equation describes a nondegenerate conic or the empty set, as follows.

If $B^2 - 4AC < 0$ , it is an ellipse (a circle when $B = 0$ and $A = C$ ) when $(A + C)\Delta < 0$ , and empty when $(A + C)\Delta > 0$ .
If $B^2 - 4AC = 0$ , it is a parabola.
If $B^2 - 4AC > 0$ , it is a hyperbola.

If $\Delta = 0$ , the equation is degenerate, and its solution set is a point, one line, two lines, or empty.

In plain words The sign of $B^2 - 4AC$ tells the type, the same way $e < 1$ , $e = 1$ and $e > 1$ do. The determinant $\Delta$ checks that the curve has not collapsed into lines or a point. For the ellipse type, the sign of $(A + C)\Delta$ then tells a real ellipse apart from an equation with no solutions, such as $x^2 + y^2 + 1 = 0$ .

Example For $5x^2 - 6xy + 5y^2 - 8 = 0$ , $B^2 - 4AC = 36 - 100 = -64 < 0$ and $\Delta = -8(25 - 9) = -128$ . Since $(A + C)\Delta = 10 \cdot (-128) < 0$ , it is a real ellipse, as found above. On the other hand $x^2 - y^2 = 0$ has $B^2 - 4AC = 4 > 0$ but $\Delta = 0$ , and indeed it is the pair of lines $y = x$ and $y = -x$ , not a hyperbola.

Check yourself

Classify $x^2 + 4xy + y^2 = 3$ .

Show the answer

Here $A = C = 1$ , $B = 4$ and $F = -3$ . The discriminant is $B^2 - 4AC = 16 - 4 = 12 > 0$ and $\Delta = -3(1 - 4) = 9 \ne 0$ , so it is a hyperbola.

Complete the square in $y^2 + 4y - 4x + 8 = 0$ and identify the curve.

Show the answer

Since $y^2 + 4y = (y + 2)^2 - 4$ , the equation becomes $(y + 2)^2 - 4 = 4x - 8$ , so $(y + 2)^2 = 4(x - 1)$ . It is a parabola with vertex $(1, -2)$ opening to the right, with $p = 1$ .

What rotation angle removes the cross term from $x^2 + 2\sqrt{3}\,xy + 3y^2 + 2x = 0$ ?

Show the answer

The angle satisfies

\cot 2\theta = \frac{1 - 3}{2\sqrt{3}} = -\frac{1}{\sqrt{3}},

and with $0 < 2\theta < \pi$ this gives $2\theta = \frac{2\pi}{3}$ , so $\theta = \frac{\pi}{3}$ .

General Conics

Key terms

General quadratic equation

Shifted conics

Completing the square

Rotation of axes

Removing the cross term

Discriminant classification

Check yourself

How this lesson connects

Builds on