Conic Sections in General Form

Quick answer

Every circle, ellipse, parabola and hyperbola with axes parallel to the coordinate axes can be written as Ax² + Cy² + Dx + Ey + F = 0. The two squared coefficients name the curve: equal gives a circle, unequal with the same sign an ellipse, opposite signs a hyperbola, and exactly one of them zero a parabola. Completing the square in x and in y rewrites the equation in terms of x − h and y − k, which locates the center or vertex and puts the curve in standard form.

What you'll learn

Classify a conic from the coefficients of x² and y²
Complete the square in x and in y to reach standard form
Read the center, radius, vertices and axes from standard form
Recognize the degenerate cases

One family, four curves

Slice a cone with a plane and the edge of the cut is a circle, an ellipse, a parabola or a hyperbola. In coordinates, all four, with axes parallel to the coordinate axes, are the same kind of equation:

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

The two squared coefficients decide which curve appears.

$A$ and $C$	Curve
equal and nonzero	circle
same sign, not equal	ellipse
opposite signs	hyperbola
exactly one is $0$	parabola

So $3x^2 + 3y^2 - 12x = 0$ is a circle, $x^2 + 4y^2 = 16$ an ellipse, $2x^2 - 5y^2 + 4 = 0$ a hyperbola, and $y = x^2 - 4x$ a parabola, since it has no $y^2$ term.

Completing the square

Classifying takes a glance. Locating the curve takes completing the square, once in $x$ and once in $y$ :

Group the $x$ terms and the $y$ terms, and move the constant to the right.
Factor the coefficient out of each group.
Complete each square, adding the matching amount to the right side.
Divide to reach standard form.

Why completing the square finds the center

Expanding $(x - h)^2 = x^2 - 2hx + h^2$ shows that a shift left or right puts a linear term in $x$ into the equation, with coefficient $-2h$ . Reading that backward, a linear term is the trace of a shift, and completing the square recovers it: $h = -\tfrac{D}{2A}$ . The same holds for $y$ . A conic in general form is a standard conic moved off the origin, and completing the square rewrites it in $x - h$ and $y - k$ , which names the move.

Once the equation reads

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

everything is where it was for a centered curve, shifted by $(h, k)$ .

ellipse centered at (1, −2)

The ellipse (x − 1)²/9 + (y + 2)²/4 = 1

Worked examples

Example 2 — An ellipse

Put $4x^2 + 9y^2 - 8x + 36y + 4 = 0$ in standard form.

Group, then factor out $4$ and $9$ :

4\left(x^2 - 2x\right) + 9\left(y^2 + 4y\right) = -4

Completing the squares inside the brackets adds $4(1) = 4$ and $9(4) = 36$ to the left, so add both to the right:

4(x - 1)^2 + 9(y + 2)^2 = 36 \quad\Longrightarrow\quad \frac{(x - 1)^2}{9} + \frac{(y + 2)^2}{4} = 1

An ellipse with center $(1, -2)$ , half-axes $a = 3$ across and $b = 2$ up, and $c = \sqrt{9 - 4} = \sqrt{5}$ , so its foci are $\sqrt{5}$ either side of the center along the horizontal axis.

Common mistakes

Practice problems

Classify $2x^2 + 2y^2 - 8x + 12y - 6 = 0$ .

Answer

A circle

Full solution

The coefficients of $x^2$ and $y^2$ are both $2$ .
Put the equation in exercise 1 in standard form, and give its center and radius.

Answer

$(x - 2)^2 + (y + 3)^2 = 16$ ; center $(2, -3)$ , radius $4$

Full solution

Divide by $2$ : $x^2 + y^2 - 4x + 6y - 3 = 0$ . Completing both squares gives $(x - 2)^2 - 4 + (y + 3)^2 - 9 = 3$ .
Classify $9x^2 + 4y^2 + 18x - 16y - 11 = 0$ , then put it in standard form.

Answer

An ellipse: $\tfrac{(x + 1)^2}{4} + \tfrac{(y - 2)^2}{9} = 1$

Full solution

$9\left(x^2 + 2x\right) + 4\left(y^2 - 4y\right) = 11$ . Completing the squares adds $9$ and $16$ to the right: $9(x + 1)^2 + 4(y - 2)^2 = 36$ . Divide by $36$ .
Give the center of the ellipse in exercise 3, and say which axis is longer.

Answer

Center $(-1, 2)$ ; the vertical axis is longer

Full solution

Under $(y - 2)^2$ sits $9$ , larger than the $4$ under $(x + 1)^2$ , so the curve reaches $3$ up and down but only $2$ left and right.
Put $x^2 + 6x - 4y + 13 = 0$ in standard form, and give its vertex.

Answer

$(x + 3)^2 = 4(y - 1)$ ; vertex $(-3, 1)$

Full solution

$x^2 + 6x = 4y - 13$ , so $(x + 3)^2 = 4y - 13 + 9 = 4(y - 1)$ . It is a parabola opening upward.
Find the focus of the parabola in exercise 5.

Answer

$(-3, 2)$

Full solution

Matching $(x - h)^2 = 4p(y - k)$ gives $4p = 4$ , so $p = 1$ . The focus sits $1$ unit above the vertex.
Classify $4y^2 - x^2 + 8y + 4x - 16 = 0$ , then put it in standard form.

Answer

A hyperbola: $\tfrac{(y + 1)^2}{4} - \tfrac{(x - 2)^2}{16} = 1$

Full solution

$4\left(y^2 + 2y\right) - \left(x^2 - 4x\right) = 16$ . Completing the squares adds $4$ to the left from the first group and subtracts $4$ from the second: $4(y + 1)^2 - (x - 2)^2 = 16$ . Divide by $16$ .
Which way does the hyperbola in exercise 7 open, and where is its center?

Answer

Up and down, centered at $(2, -1)$

Full solution

The positive square is the one in $y$ , so the branches open along the vertical axis from the center $(2, -1)$ .
What does $x^2 - y^2 = 0$ graph?

Answer

The pair of lines $y = x$ and $y = -x$

Full solution

Factor: $(x - y)(x + y) = 0$ , so either $y = x$ or $y = -x$ . It is a degenerate hyperbola, the cut a plane makes straight through the cone’s tip.
A student sees $4x^2 - 9y^2 + 8x + 36y - 68 = 0$ , notices two squared terms, and calls it an ellipse. What went wrong?

Hint

Compare the signs of the two coefficients.

Answer

The signs are opposite, so it is a hyperbola.

Full solution

An ellipse needs $A$ and $C$ to share a sign, so that the curve closes up. Here $A = 4$ and $C = -9$ , and completing the squares gives $4(x + 1)^2 - 9(y - 2)^2 = 36$ , a hyperbola centered at $(-1, 2)$ .

Frequently asked questions

What is the general form of a conic section?

Ax² + Cy² + Dx + Ey + F = 0, for a conic whose axes are parallel to the coordinate axes. An xy term would mean the curve is rotated.

How do you tell which conic an equation describes?

Compare A and C. Equal and nonzero gives a circle, unequal with the same sign an ellipse, opposite signs a hyperbola, and exactly one zero a parabola.

Why complete the square for a conic?

It collects the x terms into (x − h)² and the y terms into (y − k)², which reveals the center or vertex and puts the equation in standard form.

What is a degenerate conic?

An equation in the conic family whose graph is a point, a line, a pair of lines, or nothing at all. x² + y² = 0 is the single point (0, 0).

Does a parabola have both squared terms?

No. A parabola squares exactly one variable. If both x² and y² appear, the curve is a circle, an ellipse or a hyperbola.

What to learn next

Parametric EquationsDescribe a curve by giving x and y as functions of a parameter t: plotting and direction, eliminating the parameter, line segments and projectile motion.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSG.GPE.A.1Expressing Geometric Properties with EquationsDerive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation.
CCSS.MATH.CONTENT.HSG.GPE.A.3Expressing Geometric Properties with Equations(+) Derive the equations of ellipses and hyperbolas given the foci, using the fact that the sum or difference of distances from the foci is constant.

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