ACT Math · Geometry · free lesson

Conic Sections: Circles, Ellipses and Hyperbolas

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

Curves with an equation in $x^{2}$ and $y^{2}$

A line has an equation in $x$ and $y$ . When $x$ and $y$ are squared, the graph bends into a closed or open curve called a conic section: it is the curve where a flat plane cuts a cone. The ACT asks about three of them, the circle, the ellipse and the hyperbola. (The fourth, the parabola, has its own topic.) For each one you read a few facts straight from its equation: the center, the lengths of its axes, the points where it turns.

The circle

A circle is every point at one distance $r$ , the radius, from a center $(h, k)$ . By the Pythagorean theorem a point $(x, y)$ is at that distance when $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Take the center $(3, -2)$ and the radius 5. The engine writes the equation:

\begin{aligned} C &= \operatorname{circle}(\operatorname{point}(h, k), r) \\ &= \operatorname{circle}(\operatorname{point}(3, -2), 5) \\ &= (x - 3)^{2} + (y + 2)^{2} = 25 \end{aligned}

Watch the circle trace itself: the moving point stays 5 from the center the whole way round.

Read the equation backward with care. The signs in the brackets are the opposite of the center's coordinates: $y + 2$ means $k = -2$ . The right side is $r^{2}$ , so 25 means a radius of 5, not 25.

When the squares have been multiplied out, as in $x^2 + y^2 - 4x + 6y = 12$ , complete the square for $x$ and for $y$ : halve each coefficient, square it, and add it to both sides.

\begin{aligned} p &= (-\frac{4}{2})^{2} \\ &= 4 \\ q &= (\frac{6}{2})^{2} \\ &= 9 \\ S &= 12 + p + q \\ &= 12 + 4 + 9 \\ &= 25 \\ r &= \sqrt{S} \\ &= \sqrt{25} \\ &= 5 \end{aligned}

The center's coordinates are the halves with their signs changed:

\begin{aligned} h &= \frac{4}{2} \\ &= 2 \\ k &= -\frac{6}{2} \\ &= -3 \\ C &= \operatorname{circleFromGeneral}(-\frac{4}{2}, \frac{6}{2}, -12) \\ &= (x - 2)^{2} + (y + 3)^{2} = 25 \end{aligned}

So the center is $({2}, {-3})$ and the radius is $5$ , not the square root of 12. The last line is the engine's own conversion, and it agrees.

The ellipse

An ellipse is a stretched circle. With center $(h, k)$ its equation is

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

Here $a$ is the distance from the center to the curve along the horizontal and $b$ the distance along the vertical. The longer of the two directions is the major axis, the shorter the minor axis. The ends of the major axis are the vertices. Take $a = 5$ and $b = 3$ with center at the origin:

\begin{aligned} E &= \operatorname{ellipse}(\operatorname{point}(0, 0), a, b) \\ &= \operatorname{ellipse}(\operatorname{point}(0, 0), 5, 3) \\ &= \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 \\ V &= \operatorname{vertices}(E) \\ &= \left[ \left(5,\ 0\right),\ \left(-5,\ 0\right) \right] \\ W &= \operatorname{coVertices}(E) \\ &= \left[ \left(0,\ 3\right),\ \left(0,\ -3\right) \right] \\ major &= 2a \\ &= 2 \times 5 \\ &= 10 \\ minor &= 2b \\ &= 2 \times 3 \\ &= 6 \end{aligned}

The major axis runs between the two vertices, so its length is $2a$ , $10$ ; the minor axis has length $2b$ , $6$ . Drag the sliders. When $a$ and $b$ are equal the ellipse is a circle, and when $b$ is the larger the major axis turns upright.

Warning.

The numbers under $x^{2}$ and $y^{2}$ are $a^{2}$ and $b^{2}$ , not $a$ and $b$ . In $\frac{x^2}{25} + \frac{y^2}{9} = 1$ the ellipse reaches $5$ to each side, not $25$ , and the major axis is $10$ long.

The hyperbola

Change the plus sign to a minus and the curve breaks into two separate branches that open away from each other:

\begin{aligned} \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} &= 1 \end{aligned}

The branches turn at the vertices, $a$ to the left and right of the center. Far from the center each branch runs closer and closer to two lines through the center, the asymptotes, with slopes $\frac{b}{a}$ and $-\frac{b}{a}$ . Take $a = 4$ and $b = 3$ , center at the origin:

\begin{aligned} H &= \operatorname{hyperbola}(\operatorname{point}(0, 0), a, b) \\ &= \operatorname{hyperbola}(\operatorname{point}(0, 0), 4, 3) \\ &= \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1 \\ V &= \operatorname{vertices}(H) \\ &= \left[ \left(4,\ 0\right),\ \left(-4,\ 0\right) \right] \\ m &= \frac{b}{a} \\ &= \frac{3}{4} \\ L &= \operatorname{asymptotes}(H) \\ &= \left[ -3x + 4y = 0,\ 3x + 4y = 0 \right] \end{aligned}

The asymptotes are $y = {\frac{3}{4}}x$ and $y = -{\frac{3}{4}}x$ ; the engine's last line gives the same two lines with everything on one side. They are the diagonals of the gray box that reaches $a$ across and $b$ up from the center. Watch the two branches trace themselves along those lines.

When the $y$ term is the positive one, as in $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ , the branches open up and down instead, the vertices are $a$ above and below the center, and the asymptotes have slopes $\pm\frac{a}{b}$ .

Take $a = 3$ and $b = 4$ with the $y$ term first:

\begin{aligned} K &= \operatorname{hyperbola}(\operatorname{point}(0, 0), p, q, \text{vertical}) \\ &= \frac{y^{2}}{9} - \frac{x^{2}}{16} = 1 \\ W &= \operatorname{vertices}(K) \\ &= \left[ \left(0,\ 3\right),\ \left(0,\ -3\right) \right] \\ A &= \operatorname{asymptotes}(K) \\ &= \left[ 3x - 4y = 0,\ 3x + 4y = 0 \right] \end{aligned}

The vertices are on the $y$ -axis, and the asymptotes are $y = \pm{{p/q}}x$ .

Which conic is it?

In an equation $Ax^{2} + Cy^{2} + Dx + Ey + F = 0$ the two squared terms decide. The engine names each of three equations:

\begin{aligned} first &= \operatorname{conicType}(\operatorname{conic}(1, 0, 1, -6, 8, -11)) \\ &= \text{circle} \\ second &= \operatorname{conicType}(\operatorname{conic}(4, 0, 9, -16, 54, 61)) \\ &= \text{ellipse} \\ third &= \operatorname{conicType}(\operatorname{conic}(9, 0, -4, 0, 0, -36)) \\ &= \text{hyperbola} \end{aligned}

The $x^{2}$ and $y^{2}$ terms	The curve	Example
same sign, same coefficient	circle	$x^2 + y^2 - 6x + 8y - 11 = 0$
same sign, different coefficients	ellipse	$4x^2 + 9y^2 - 16x + 54y + 61 = 0$
opposite signs	hyperbola	$9x^2 - 4y^2 - 36 = 0$
only one of them	parabola	$y = x^2 - 4x$

Note.

On the ACT, conic questions are medium to hard. Expect to read the center and radius of a circle (with the signs changed and the square root taken), write a circle's equation from its center and a point on it, complete the square, count the points where a circle meets a line or two conditions on distance, read the axes of an ellipse, find a hyperbola's asymptotes, and name the conic an equation describes.

Worked examples

Each example is one of the kinds of question the ACT asks on circles, ellipses and hyperbolas, from easy to hard. The numbers in the working are found by the engine.

Example.

1. Center and radius. A circle in the standard $(x, y)$ coordinate plane has the equation $(x + 4)^{2} + (y - 1)^{2} = 49$ . What are its center and its radius?

Compare with $(x - h)^{2} + (y - k)^{2} = r^{2}$ . The bracket $x + 4$ is $x - (-4)$ , so $h = -4$ ; the bracket $y - 1$ gives $k = 1$ . The right side is $r^{2}$ , so take its square root:

\begin{aligned} r &= \sqrt{S} \\ &= \sqrt{49} \\ &= 7 \\ C &= \operatorname{circle}(\operatorname{point}(h, k), r) \\ &= \operatorname{circle}(\operatorname{point}(-4, 1), 7) \\ &= (x + 4)^{2} + (y - 1)^{2} = 49 \end{aligned}

The center is $({-4}, {1})$ and the radius is $7$ . Building the circle back from them gives the equation in the question, which is the check. The usual slips are the center $(4, -1)$ and the radius 49.

Example.

2. The equation from the center and a point. A circle has center $(2, -3)$ and passes through the point $(5, 1)$ . What is its equation?

The radius is the distance from the center to the point. The equation needs $r^{2}$ , so the square of that distance is all you need: the differences of the coordinates, squared and added.

\begin{aligned} S &= (x - h)^{2} + (y - k)^{2} \\ &= (5 - 2)^{2} + (1 - (-3))^{2} \\ &= 25 \\ r &= \sqrt{S} \\ &= \sqrt{25} \\ &= 5 \\ C &= \operatorname{circle}(\operatorname{point}(h, k), r) \\ &= \operatorname{circle}(\operatorname{point}(2, -3), 5) \\ &= (x - 2)^{2} + (y + 3)^{2} = 25 \end{aligned}

The radius is $5$ and the equation is $(x - 2)^{2} + (y + 3)^{2} = 25$ . Check: the point $(5, 1)$ is on the circle because it gave the right side in the first line.

Example.

3. Completing the square. What are the center and the radius of the circle $x^2 + y^2 + 10x - 4y - 7 = 0$ ?

Move the number to the right side: $x^2 + 10x + y^2 - 4y = 7$ . Halve the coefficient of $x$ and square it; do the same for $y$ ; add both squares to both sides. The right side becomes $r^{2}$ .

\begin{aligned} p &= (\frac{10}{2})^{2} \\ &= 25 \\ q &= (-\frac{4}{2})^{2} \\ &= 4 \\ S &= 7 + p + q \\ &= 7 + 25 + 4 \\ &= 36 \\ r &= \sqrt{S} \\ &= \sqrt{36} \\ &= 6 \end{aligned}

The left side is now $(x + 5)^{2} + (y - 2)^{2}$ , so the center has the halves with their signs changed:

\begin{aligned} h &= -\frac{10}{2} \\ &= -5 \\ k &= \frac{4}{2} \\ &= 2 \\ C &= \operatorname{circleFromGeneral}(\frac{10}{2}, -\frac{4}{2}, -7) \\ &= (x + 5)^{2} + (y - 2)^{2} = 36 \end{aligned}

The center is $({-5}, {2})$ and the radius is $6$ ; the engine's own conversion on the last line agrees. Taking $\sqrt{7}$ as the radius, without adding the two squares, is the trap.

Example.

4. The axes of an ellipse. For the ellipse $\frac{(x - 1)^2}{36} + \frac{(y + 2)^2}{16} = 1$ , find the center, the length of the major axis and its endpoints.

The center is $(1, -2)$ , with the signs changed as for a circle. The denominators are the squares of the half-axes, so take square roots: $a$ across and $b$ up and down.

\begin{aligned} a &= \sqrt{36} \\ &= 6 \\ b &= \sqrt{16} \\ &= 4 \\ E &= \operatorname{ellipse}(\operatorname{point}(h, k), a, b) \\ &= \operatorname{ellipse}(\operatorname{point}(1, -2), 6, 4) \\ &= \frac{(x - 1)^{2}}{36} + \frac{(y + 2)^{2}}{16} = 1 \\ major &= 2a \\ &= 2 \times 6 \\ &= 12 \\ minor &= 2b \\ &= 2 \times 4 \\ &= 8 \\ V &= \operatorname{vertices}(E) \\ &= \left[ \left(7,\ -2\right),\ \left(-5,\ -2\right) \right] \\ W &= \operatorname{coVertices}(E) \\ &= \left[ \left(1,\ 2\right),\ \left(1,\ -6\right) \right] \end{aligned}

The larger half-axis, $6$ , is across, so the major axis is horizontal and $12$ long. Its endpoints, the vertices, are $6$ to the left and right of the center. The minor axis is $8$ long, between the co-vertices. Reading 36 as the half-axis instead of its square root is the usual slip.

Example.

5. A hyperbola's vertices and asymptotes. For the hyperbola $\frac{x^2}{9} - \frac{y^2}{25} = 1$ , find the vertices and the equations of the asymptotes.

The $x$ term is the positive one, so the branches open left and right. The vertices are $a$ from the center along the $x$ -axis, and the asymptotes through the center have slopes $\pm\frac{b}{a}$ .

\begin{aligned} a &= \sqrt{9} \\ &= 3 \\ b &= \sqrt{25} \\ &= 5 \\ H &= \operatorname{hyperbola}(\operatorname{point}(0, 0), a, b) \\ &= \operatorname{hyperbola}(\operatorname{point}(0, 0), 3, 5) \\ &= \frac{x^{2}}{9} - \frac{y^{2}}{25} = 1 \\ V &= \operatorname{vertices}(H) \\ &= \left[ \left(3,\ 0\right),\ \left(-3,\ 0\right) \right] \\ m &= \frac{b}{a} \\ &= \frac{5}{3} \end{aligned}

The vertices are $({3}, 0)$ and $({-3}, 0)$ , and the asymptotes are $y = {\frac{5}{3}}x$ and $y = -{\frac{5}{3}}x$ . The slip is to turn the fraction over and write $\pm\frac{a}{b}$ : the slope is rise over run, $b$ up for every $a$ across.

Example.

6. Counting points. How many points in the standard $(x, y)$ coordinate plane are 5 units from the origin and 3 units from the $x$ -axis?

The points 5 units from the origin make the circle $x^{2} + y^{2} = 25$ . The points 3 units from the $x$ -axis make two lines, $y = 3$ above it and $y = -3$ below it. Count where the lines cross the circle. On the line $y = 3$ :

\begin{aligned} x &= \sqrt{r^{2} - d^{2}} \\ &= \sqrt{5^{2} - 3^{2}} \\ &= 4 \end{aligned}

So the line $y = 3$ meets the circle at $x = 4$ and $x = -4$ , two points. The line $y = -3$ does the same by symmetry, two more. There are 4 points in all: $({4}, 3)$ , $({-4}, 3)$ , $({4}, -3)$ and $({-4}, -3)$ . Answering 2 forgets the line below the axis.

Practice

Answer ACT-style questions on conic sections: circles, ellipses and hyperbolas. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise conic sections: circles, ellipses and hyperbolas with new numbers

Conic Sections: Circles, Ellipses and Hyperbolas

The idea

Curves with an equation in x2x^{2}x2 and y2y^{2}y2

The circle

The ellipse

The hyperbola

Which conic is it?

Worked examples

Worked examples

Practice

Curves with an equation in $x^{2}$ and $y^{2}$