Chapter 10.4, Problem 22AYU

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

Center at $\left(0,0\right)$; focus at $\left(-3,0\right)$; vertex at $\left(2,0\right)$

To determine

a & b

To find: An equation of the hyperbola described.

Answer to Problem 22AYU

Solution:

$\frac{x^2}{4}-\frac{y^2}{5}=1$

Explanation of Solution

Given:

Center at $\left(0,0\right)$; focus at $\left(-3,0\right)$; vertex at $\left(2,\text{ 0}\right)$.

Formula used:

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

Calculation:

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

Focus: $\left(-3,0\right)=\left(\pm c,0\right)$; $c=3$.

Vertex: $\left(2,0\right)=\left(a,0\right);a=2$.

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

Hence, the equation of the hyperbola is,

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

To find b;

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

$b²=3²-2²=9-4=5;b=\surd 5$

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

$\frac{x^2}{2²}-\frac{y^2}{\surd 5^2}=1$

$\frac{x^2}{4}-\frac{y^2}{5}=1$

Asymptotes are $y=\pm \frac{\sqrt{5}}{2}$.

c.

To graph:

The equation $\frac{x^2}{4}-\frac{y^2}{5}=1$.

Solution:

Given:

Center at $\left(0,0\right)$; focus at $\left(-3,0\right)$; vertex at $\left(2,0\right)$.

Graph:

The graph of the equation of $\frac{x^2}{4}-\frac{y^2}{5}=1$ is plotted.

Using the given information center at $\left(0,0\right)$; focus at $\left(-3,0\right)$; vertex at $\left(2,\text{ 0}\right)$, we find $a=2$, $b=\surd 5$. We have formed the equation and plotted the graph for $\frac{x^2}{4}-\frac{y^2}{5}=1$. The transversal axis is $x\text{-axis}$.