Conic Sections — AP® Precalculus Notes

Beyond the exam. The AP® Precalculus exam does not test this unit. Many classes teach it, and it prepares you for calculus.

A double cone, two ice-cream cones point to point, can be sliced by a plane in different ways. A cut straight across gives a circle; a tilted cut gives an ellipse; a cut parallel to the side gives a parabola; a steep cut through both cones gives the two branches of a hyperbola. These curves are the conic sections. Each is an implicitly defined curve, with an equation in $x$ and $y$ .

Three graphs. Left: the parabola y = 0.5 x squared with its vertex at the origin. Middle: the ellipse x squared over 9 plus y squared over 4 = 1, centered at the origin, with vertices (3, 0) and (-3, 0) and the ends of the short axis at (0, 2) and (0, -2). Right: the hyperbola x squared minus y squared over 2.25 = 1, with vertices (1, 0) and (-1, 0), two branches opening left and right, and dashed asymptotes y = 1.5x and y = -1.5x. — **Figure 1.** The three conic sections in standard position. LevlPrep original diagram.

Parabolas

You know $y = a(x - h)^2 + k$ , a parabola with vertex $(h, k)$ that opens up if $a > 0$ and down if $a < 0$ . Swap the roles of $x$ and $y$ and you get a parabola that opens sideways:

x - h = a(y - k)^2.

It has vertex $(h, k)$ and opens right if $a > 0$ , left if $a < 0$ . A sideways parabola is not the graph of a function of $x$ , since most $x$ -values have two $y$ -values.

Ellipses

An ellipse with center $(h, k)$ has the standard form

\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1, \qquad a > 0, \; b > 0.

It reaches $a$ units left and right of the center and $b$ units up and down. The farthest points along the longer direction are the vertices; the longer axis has length $2a$ or $2b$ , whichever is larger. When $a = b$ the ellipse is a circle of radius $a$ .

Worked example 1. Describe $\frac{(x - 1)^2}{16} + \frac{(y + 2)^2}{4}$ $= 1$ .

Center: $x - 1$ and $y + 2 = y - (-2)$ give $(1, -2)$ .

Distances: $a^2 = 16$ so $a = 4$ across; $b^2 = 4$ so $b = 2$ up and down. The ellipse is wider than tall.

Vertices: $(1 \pm 4, -2)$ , that is $(5, -2)$ and $(-3, -2)$ ; the long axis has length $8$ . The top and bottom points are $(1, 0)$ and $(1, -4)$ .

Hyperbolas

Change the plus sign to a minus and the curve breaks into two branches:

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

This hyperbola opens left and right. Its vertices are $(h \pm a, k)$ , and far from the center each branch gets closer and closer to the asymptotes $y - k = \pm\frac{b}{a}(x - h)$ . If the $y$ term comes first, $\frac{(y - k)^2}{b^2} - \frac{(x - h)^2}{a^2}$ $= 1$ , it opens up and down with vertices $(h, k \pm b)$ and the same asymptotes.

Worked example 2. Describe $\frac{x^2}{4} - \frac{y^2}{9} = 1$ .

Center $(0, 0)$ , $a = 2$ , $b = 3$ ; the $x$ term is positive, so it opens left and right. Vertices $(2, 0)$ and $(-2, 0)$ . Asymptotes $y = \pm\frac{3}{2}x$ .

It never meets the $y$ -axis: $x = 0$ gives $-\frac{y^2}{9} = 1$ , so $y^2 = -9$ , which has no real solution.

Which conic is it?

Squared terms	Conic	Example
only one variable squared	parabola	$x = 2y^2 - 3$
both squared, same sign	ellipse (circle if the coefficients match)	$4x^2 + 9y^2 = 36$
both squared, opposite signs	hyperbola	$y^2 - 4x^2 = 16$

To read the details, rewrite in standard form. Divide so the right side is $1$ : $4x^2 + 9y^2 = 36$ becomes $\frac{x^2}{9} + \frac{y^2}{4} = 1$ . If there are linear terms, complete the square.

Worked example 3. Identify $x^2 + y^2 - 6x + 4y - 3$ $= 0$ .

Group and move the constant: $(x^2 - 6x) + (y^2 + 4y)$ $= 3$ .

Complete each square: add $9$ and $4$ to both sides, $(x - 3)^2 + (y + 2)^2 = 16$ .

A circle with center $(3, -2)$ and radius $4$ .