# Hyperbolas

### Parts of a Hyperbola

A hyperbola is a conic section defined by two fixed points. The difference of the distances between these two fixed points and any point on the hyperbola is constant.

A hyperbola is the set of points such that the difference of the distances from two fixed points called the foci remains the same. A hyperbola is formed when a plane intersects, or crosses, both halves of a double cone without passing through its apex. The graph of a hyperbola has two unconnected branches that are mirror images of each other.

The properties of a hyperbola include:

• A hyperbola has two axes of symmetry. The transverse axis is the line that passes through the foci of a hyperbola. The conjugate axis is the line perpendicular to the transverse axis that intersects it at the center of the hyperbola.
• A vertex of a hyperbola is one of two points where the transverse axis intersects the hyperbola.
• A hyperbola has two asymptotes that cross at the center of the hyperbola. An asymptote is a line that a graph approaches as one of the variables approaches infinity or negative infinity.
• The fundamental rectangle is a rectangle used as a guide to graph a hyperbola. The vertices of a hyperbola are the midpoints of two sides of the fundamental rectangle.
• The diagonals of the fundamental rectangle, or the segments connecting opposite vertices, lie on the asymptotes of the hyperbola.
The equation of a hyperbola can be written in terms of its center and vertices.

### Equation of a Hyperbola

 Horizontal Transverse Axis Centered at the Origin Vertical Transverse Axis Centered at the Origin The equation is: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ The equation is: $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ The coordinates of the vertices are $(-a,0)$ and $(a,0)$. The coordinates of the vertices are $(0, -a)$ and $(0,a)$. The coordinates of the foci are $(c,0)$ and $(-c,0)$, where: $a^2+b^2=c^2$ The coordinates of the foci are $(0,c)$ and $(0,-c)$ , where: $a^2+b^2=c^2$

The standard form for an equation of a hyperbola with center $(h, k)$ is written using $(x-h)$ in place of $x$ and $(y-k)$ in place of $y$.

• Horizontal transverse axis:
$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$
• Vertical transverse axis:
$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$

### Graphing Hyperbolas

A hyperbola with a given equation can be graphed in the coordinate plane by locating the center and vertices and sketching its asymptotes and fundamental rectangle.
A hyperbola can have its center at the origin or elsewhere on the coordinate plane, and its transverse axis, or the axis that passes through its foci and vertices, can be horizontal or vertical.

### Hyperbolas in the Coordinate Plane

Horizontal Transverse Axis Vertical Transverse Axis
Equation
$\frac{x^2}{25}-\frac{y^2}{9}=1$
$\frac{y^2}{4}-\frac{x^2}{16}=1$
Center
$(0, 0)$
$(0, 0)$
Vertices $(-5, 0)$ and $(5, 0)$ $(0, -2)$ and $(0, 2)$
Transverse axis
$y=0$
$x=0$
Conjugate axis
$x=0$
$y=0$
Graph

To graph a hyperbola given its equation in standard form:

1. Use the equation to identify the center of the hyperbola.
2. Use the values of $a$ and $b$ to draw the fundamental rectangle. The length of the rectangle corresponds to the distance between the vertices and is equal to $2a$, and the width of the rectangle is equal to $2b$.
3. Draw diagonal lines through opposite corners of the fundamental rectangle to show the asymptotes.
4. Locate the vertices.
5. Plot additional points on the hyperbola as needed, and sketch two curves through the vertices that approach the asymptotes.

As with circles and ellipses, sometimes, the equation of a hyperbola may need to be rewritten in standard form by completing the square.

Step-By-Step Example
Graphing a Hyperbola in Standard Form
Graph the hyperbola:
$\frac{(y-2)^2}{9}-\frac{(x-3)^2}{9}=1$
Step 1
Compare the equation to the standard form for the equation of a hyperbola. The $x^2$-term is subtracted from the $y^2$-term, so the standard form is:
$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$
This means that the transverse axis is vertical and the conjugate axis is horizontal.
$\frac{(y-2)^2}{9}-\frac{(x-3)^2}{9}=1$
So, $h=3$, $k=2$, $a^2=9$, and $b^2=9$.

The transverse axis is the vertical line $x=h$, or $x=3$.
The conjugate axis is the horizontal line $y=k$, or $y=2$.

Step 2

Determine the center of the hyperbola and the lengths of $a$ and $b$.

Since $h=3$ and $k=2$, the center is $(3,2)$.

For the length of $a$:
\begin{aligned}a^{2}&=9\\a&=\sqrt{9}\\&=3\end{aligned}
For the length of $b$:
\begin{aligned}b^{2}&=9\\b&=\sqrt{9}\\&=3\end{aligned}
Plot the center, and graph the transverse and conjugate axes. The transverse axis is vertical, and the conjugate axis is horizontal. Use the center and the values of $a$ and $b$ to draw the fundamental rectangle. Since $a=b$, the fundamental rectangle of the hyperbola is a square.
Step 3
Draw the asymptotes through the corners of the fundamental rectangle.
Step 4

Calculate the vertices. The vertices are both 3 units from the center. The transverse axis is vertical, so the vertices are above and below the center.

Vertices:
$\begin{gathered}(3, 2+3)=(3,5) \\ (3, 2-3)=(3,-1)\end{gathered}$
Notice that the vertices are the points where the transverse axis intersects the fundamental rectangle.
Step 5

Identify the foci.

To locate the foci, use the equation:
$a^2+b^2=c^2$
Determine the value of $c$ by using the values $a=3$ and $b=3$ from Step 2.
\begin{aligned}a^2+b^2&=c^2\\3^2+3^2&=c^2\\9+9&=c^2\\18&=c^2\\\sqrt{18}&=c\\3\sqrt{2}&=c\end{aligned}
The transverse axis is vertical. So, the foci are $c$ units above and below the center: $\left(3,2+3\sqrt{2}\right)$ and $\left(3,2-3\sqrt{2}\right)$, or about $(3,6.2)$ and $(3,-2.2)$.
Solution

By Hand: Plot the vertices and foci. Draw a smooth curve through the vertex $(3,5)$ that approaches both asymptotes above the fundamental rectangle. Draw another smooth curve through the vertex $(3,-1)$ that approaches both asymptotes below the fundamental rectangle.

On a Graphing Calculator: Solve for $y$.
\begin{aligned}\frac{(y-2)^2}{9}-\frac{(x-3)^2}{9}&=1\\y&=\pm\sqrt{9-(x-3)^2}+2\end{aligned}
Graph both branches of the hyperbola:
\begin{aligned}Y_1&=\sqrt{9-(x-3)^2}+2 \\ Y_2&=-\sqrt{9-(x-3)^2}+2\end{aligned}
Step-By-Step Example
Graphing a Hyperbola by Completing the Square
Graph the hyperbola:
$x^2-4y^2-4x+40y-100=0$
Step 1
Reorder the terms so that like terms are together, and move the constant term to the right side of the equation.
\begin{aligned}x^2-4y^2-4x+40y-100&=0\\x^2-4x-4y^2+40y&=100\end{aligned}
Step 2
Group like terms. Factor the $y$-terms so that the coefficient of $y^2$ is 1.
$(x^2-4x)-4(y^2-10y)=100$
Step 3
Identify the terms that need to be added to complete the square for the $x$-terms and the $y$-terms.
$({\color{#c42126} x^2-4x+\underline{\ \ \ \ }})-4({\color{#0047af} y^2-10y+\underline{\ \ \ \ }})=100+{\color{#c42126} \underline{\ \ \ \ }}-4({\color{#0047af} \underline{\ \ \ \ }})$
The missing term for each variable is a number that will make each expression a perfect square. To find this number, divide the coefficient of the middle term by 2, and square the result. For the $x$-terms:
$\left(\frac{-4}{2}\right)^2=4$
For the $y$-terms:
$\left(\frac{-10}{2}\right)^2=25$
Step 4

Complete the square for each variable, adding the value to both sides of the equation. Then, factor the expressions on the left side into perfect squares and simplify the right side to write the equation in standard form.

When adding to both sides, remember to account for the coefficient by multiplying the amount added to the $y$-terms by –4.
\begin{aligned}({\color{#c42126} x^2-4x+\underline{\ \ \ \ }})-4({\color{#0047af} y^2-10y+\underline{\ \ \ \ }})&=100+{\color{#c42126} \underline{\ \ \ \ }}-4({\color{#0047af} \underline{\ \ \ \ }})\\({\color{#c42126} x^2-4x+4})-4({\color{#0047af} y^2-10y+25})&=100+{\color{#c42126}4}-4({\color{#0047af} 25})\\({\color{#c42126} x-2})^2-4({\color{#0047af} y-5})^2&=4\end{aligned}
Step 5
Divide both sides of the equation by 4 to write the equation in standard form, with the right side equal to 1.
\begin{aligned}(x-2)^2-4(y-5)^2&=4\\ \frac{(x-2)^2}{4}-\frac{4(y-5)^2}{4}&=\frac{4}{4}\\ \frac{(x-2)^2}{4}-\frac{(y-5)^2}{1}&=1\end{aligned}
Step 6
Identify the center of the hyperbola.
$\frac{(x-2)^2}{4}-\frac{(y-5)^2}{1}=1$
Since $h=2$ and $k=5$, the center is $(2,5)$.
Step 7
Determine the axes.
$\frac{(x-2)^2}{4}-\frac{(y-5)^2}{1}=1$
The $y^2$-term is subtracted from the $x^2$-term in the standard form of the equation. So, the transverse axis is horizontal, and the conjugate axis is vertical.

The transverse axis is the horizontal line $y=k$, or $y=5$.

The conjugate axis is the vertical line $x=h$, or $x=2$.

Step 8
Calculate the lengths of $a$ and $b$ and the coordinates of the vertices.
$\frac{(x-2)^2}{4}-\frac{(y-5)^2}{1}=1$
For the length of $a$:
\begin{aligned}a^{2}&=4\\a&=\sqrt4\\&=2\end{aligned}
For the length of $b$:
\begin{aligned}b^{2}&=1\\b&=\sqrt{1}\\&=1\end{aligned}
The coordinates of the vertices are $(h-a,k)$ and $(h+a,k)$.
The vertices are at $(0,5)$ and $(4,5)$.
Step 9
Identify the foci. To locate the foci, solve the equation $a^2+b^2=c^2$ to find the value of $c$. Use the values $a=2$ and $b=1$ from Step 8.
\begin{aligned}a^2+b^2&=c^2\\2^2+1^2&=c^2\\4+1&=c^2\\5&=c^2\\\sqrt{5}&=c\end{aligned}
The transverse axis is horizontal, so the foci are $c$ units left and right of the center: $\left(2+\sqrt{5},1\right)$ and $\left(2-\sqrt{5},1\right)$, or about $(4.2, 1)$ and $(-0.2,1)$.
Solution
• Plot the center and the vertices, and draw the transverse and conjugate axes.
• Use the values of $a$ and $b$ to draw the fundamental rectangle.
• Draw the asymptotes through the diagonals of the fundamental rectangle.
• Then sketch the two branches of the hyperbola approaching the asymptotes.