Chapter 10.4, Problem 21AYU

In Problems 19-28, find an equation for the hyperbola described. Graph the equation by hand.

Center at $\left(0,0\right)$; focus at $\left(0,-6\right)$; vertex at $\left(0,4\right)$

To determine

a & b

To find: An equation of the hyperbola described.

Answer to Problem 21AYU

Solution:

$\frac{y^2}{16}-\frac{x^2}{20}=1$

Explanation of Solution

Given:

Center at $\left(0,0\right)$; focus at $\left(0,-6\right)$; vertex at $\left(0,4\right)$.

Formula used:

Center	Transverse axis	Foci	Vertices	Equation	Asymptotes
$\left(0,0\right)$	$y\text{-axis}$	$\left(0,\pm c\right)$	$\left(0,\pm a\right)$	$\frac{y^2}{a²}-\frac{x²}{b²}=1;b^2=c^2-a^2$	$y=\frac{a}{b}x;y=-\frac{a}{b}x$

Calculation:

The center of the hyperbola is $\left(0,0\right)$;

Focus: $\left(0,-6\right)=\left(0,\pm c\right)$; $c=6$.

Vertex: $\left(0,4\right)=\left(0,\pm a\right);a=4$.

From the given data we see that, transverse axis is $y\text{-axis}$.

Hence, the equation of the hyperbola is,

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

To find b;

$b^2=c^2-a^2$

$b²=6²-4²=36-16=20;b=2\surd 5$

Substituting the values of $a$ and $b$ in the equation of the hyperbola, we get,

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

$\frac{y^2}{4²}-\frac{x²}{20}=1$

$\frac{y^2}{16}-\frac{x^2}{20}=1$

The asymptotes are $y=\pm \frac{4}{2\sqrt{5}}=\pm \frac{2}{\sqrt{5}}$.