Precalculus - Hyperbolas

Introduction

• Hyperbola is the set of points in a plane wherein the absolute value of the difference of the distances from two fixed points, the foci, is constant. The line joining both points (foci) is called the transverse axis.
• Hyperbola is formed when a cone is intersected by a plane parallel to the x-axis.

The absolute value of the differences of the distances from two fixed points is expressed as $\left|d_1-d_2\right|=2a$. 

In the hyperbola above, we have $\left|AK-BK\right|=2a$

• The general equation of a hyperbola is $A{x}^2+C{y}^2+Dx+Ey+F=0, where AC<0$.
• Its eccentricity is $>1$. The formula to find the eccentricity of the hyperbola is given by $e=\frac{c}{a}$.
• To visualize the properties of a hyperbola, study the graph below:

• Observe that the hyperbola has an auxiliary rectangle where its two opposite sides intersect the vertices of the curve.
• Two asymptotes are diagonal lines of the auxiliary rectangle whose intersection is the hyperbola's center.
• The curve of the hyperbola is symmetric to the x-axis and y-axis.
• The value of $a$, unlike the ellipse, is not necessarily greater than $b$.
• The hyperbola above has a center at $\left(0,0\right)$, foci at $A\left(5,0\right) and B\left(-5,0\right)$, vertices at $C\left(4,0\right) and D\left(-4,0\right)$, Co-vertices at $E\left(0,3\right) and F\left(0,-3\right)$. The equation of the hyperbola above is $\frac{x^2}{16}-\frac{y^2}{9}=1$.
• The asymptotes are $y=\pm \frac{3}{4}x$.

Forms and Graphs of Hyperbola

1. Center at $\left(0,0\right)$ with transverse axis along the x-axis (horizontal transverse axis)

Standard Equation is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

To find the value of c, we have: $a^2+b^2=c^2$ (center to focus distance)

The length of the latus rectum is computed using $L=\frac{2b^2}{a}$.

The asymptotes are $y=\pm \frac{b}{a}x$

The coordinates of the foci are $F_1\left(c,0\right) and F_2\left(-c,0\right)$.

The coordinates of the vertices are $V_1\left(a,0\right) and V_2\left(-a,0\right)$.

Coordinates of the co-vertices are $B_1\left(0,b\right) and B_2\left(0,-b\right)$

The length of the conjugate axis is $2b$.

Graph: 

2. Center at $\left(0,0\right)$ with a transverse axis along the y-axis (vertical transverse axis)

Standard equation is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

To find the value of c, we have $c^2=a^2+b^2$ (center to focus distance)

The length of the latus rectum is given by $L=\frac{2b^2}{a}$.

The asymptotes are $y=\pm \frac{a}{b}x$

The coordinates of the foci are $F_1\left(0,c\right) and F_2\left(0,-c\right)$.

Coordinates of the vertices are $V_1\left(0,a\right) and V_2\left(0,-a\right)$

Coordinates of the co-vertices are $B_1\left(b,0\right) and B_2\left(-b,0\right)$

The length of the conjugate axis is $2b$.

Graph:

3. Center at $\left(h,k\right)$ with a transverse axis parallel to the x-axis

Standard Equation is $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$

Center to focus distance is $c^2=a^2+b^2$.

The coordinates of the vertices are $V_1\left(h+a,k\right) and V_2\left(h-a,k\right)$.

Coordinates of the foci are $F_1\left(h+c,k\right) and F_2\left(h-c,k\right)$

Coordinates of the co-vertices (endpoints of the conjugate axis) are $B_1\left(h,k+b\right) and B_2\left(h,k-b\right)$

Asymptotes are $y-k=\pm \frac{b}{a}\left(x-h\right)$

Graph:

4. Center at $\left(h,k\right)$ with a transverse axis parallel to the y-axis.

Standard Equation is $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$

Center to focus distance is $c^2=a^2+b^2$

The coordinates of the vertices are $V_1\left(h,k+a\right) and V_2\left(h,k-a\right)$.

The coordinates of the foci are $F_1\left(h,k+c\right) and F_2\left(h,k-c\right)$.

Coordinates of the co-vertices (endpoints of the conjugate axis) are $B_1\left(h+b,k\right) and B_2\left(h-b,k\right)$.

Graph:

Solved Examples

Example 1. In the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$, locate the center, vertices, co-vertices, and foci. What is the equation of the asymptotes?

Solution:

This hyperbola has its transverse axis along the x-axis. The center is at $\left(0,0\right)$.

To solve for $a$, we have $a^2=16\to a=4$.

To find b, we have $b^2=9\to b=3$

To find the value of $c$, we use the formula $c^2=a^2+b^2$. This gives $c=5$.

Based on the computed values, we have the following coordinates:

Foci at $\left(5,0\right) and \left(-5,0\right)$

Vertices at $\left(4,0\right) and \left(-4,0\right)$.

Co-vertices at $\left(0,3\right) and \left(0,-3\right)$

The equation of the asymptotes is $y=\pm \frac{b}{a}x\to y=\pm \frac{3}{4}x$

 

Example 2. Find the equation of the hyperbola with a horizontal transverse axis, an eccentricity of 4, and a distance between the foci is $4\sqrt{3}$.

Solution:

This hyperbola has a standard equation of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

If the distance between the foci is $4\sqrt{3}$, then $2c=4\sqrt{3}\to c=2\sqrt{3}$.

To solve for a, we use eccentricity $e=\frac{c}{a}$.

$4=\frac{2\sqrt{3}}{a}\to 4a=2\sqrt{3}\to a=\frac{\sqrt{3}}{2}$

We use $c^2=a^2+b^2$ to solve for $b$. This yields $b^2=\frac{45}{4}$

Hence, the equation is $\frac{4x^2}{3}-\frac{4y^2}{45}=1 or 60x^2-4y^2=45$

 

Example 3. What is the equation of the hyperbola with vertices at $\left(0,7\right) and \left(0,-7\right)$ and one focus of $\left(0,9\right)$?

Solution:

This hyperbola has a transverse axis along the y-axis, which follows the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$.

If the vertices and one focus are $\left(0,7\right) and \left(0,-7\right)$, and $\left(0,9\right)$, then $a=7 and c=9$.

Using $c^2=a^2+b^2$ yields $b^2=32$.

Thus, the equation of the hyperbola is $\frac{y^2}{49}-\frac{x^2}{32}=1$

Example 4. Determine the center, vertices, co-vertices, and foci of the hyperbola given by $\frac{{\left(x+3\right)}^2}{36}-\frac{{\left(y-2\right)}^2}{25}=1$.

Solution:

The center is  $\left(-3,2\right)$ where $h=-3 and k=2$.

Based on the given equation, $a=6 and b=5$.

Using the formula $c^2=a^2+b^2$, then $c=\sqrt{61}$.

Foci are at $\left(-3+\sqrt{61},2\right) and \left(-3-\sqrt{61},2\right)$

Vertices at $\left(3,2\right) and \left(-9,2\right)$

Co-vertices at $\left(-3,7\right) and \left(-3,-3\right)$

 

Example 5. Find the equation of the hyperbola with center at $\left(-6,5\right)$, foci at $\left(-9,5\right) and \left(-3,5\right)$, and eccentricity of 6.

Solution:

Based on the given points, the hyperbola has a horizontal transverse axis. 

If the foci are at $\left(-9,5\right) and \left(-3,5\right)$, it follows that $-6+c=-3 and -6-c=-9$ which gives $c=3$.

Using the eccentricity $e=\frac{c}{a} or 6=\frac{c}{a}$, then $a=\frac{1}{2}$.

Using the formula $c^2=a^2+b^2$, it gives $b^2=\frac{35}{4}$.

Hence, the equation of the hyperbola is $\frac{4{\left(x+6\right)}^2}{1}-\frac{4{\left(y-5\right)}^2}{35}=1 or 140{\left(x+6\right)}^2-4{\left(y-5\right)}^2=35$

 

Example 6. Locate the center of the hyperbola $x^2-4y^2+4x+32y-96=0$.

Solution:

Use completing the square to determine the coordinates of the center.

$x^2+4x-4y^2+32y=96$

$\left(x^2+4x+4\right)-4\left(y^2-8y+16\right)=96+4+\left(-4\right)\left(16\right)$

${\left(x+2\right)}^2-4{\left(y-4\right)}^2=36$, with the center at $\left(-2,4\right)$

Cheat Sheet

Standard Forms of Hyperbola	Coordinates				Equation of the asymptotes
	Center and transverse axis	Vertices	Foci	Co-vertices (endpoints of the conjugate axis)
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$	$\left(0,0\right)$ along the x-axis	$\left(\pm a,0\right)$	$\left(\pm c,0\right)$	$\left(0,\pm b\right)$	$y=\pm \frac{b}{a}x$
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$	$\left(0,0\right)$ along the y-axis	$\left(0,\pm a\right)$	$\left(0,\pm c\right)$	$\left(\pm b, 0\right)$	$y=\pm \frac{a}{b}x$
$\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$	$\left(h,k\right)$ parallel to x-axis	$\left(h\pm a,k\right)$	$\left(h\pm c,k\right)$	$\left(h, k\pm b\right)$	$y-k=\pm \frac{b}{a}\left(x-h\right)$
$\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$	$\left(h,k\right)$ parallel to y-axis	$\left(h, k\pm a\right)$	$\left(h, k\pm c\right)$	$\left(h\pm b, k\right)$	$y-k=\pm \frac{a}{b}\left(x-h\right)$

Blunder Areas

• In the equation of the hyperbola, $a$ is not necessarily greater than or less than b, unlike in the case of the ellipse.
• Always notice the difference between the hyperbola $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1 and \frac{{\left(y-k\right)}^2}{b^2}-\frac{{\left(x-h\right)}^2}{a^2}=1$ in terms of the transverse axis and the conjugate axis.
• Be cognizant of the difference between an ellipse and a hyperbola in terms of their standard equation and general equation.