Chapter 10.4, Problem 67AYU

To determine

To find: Analyze the equation.

Answer to Problem 67AYU

The equation represents hyperbola.

Explanation of Solution

Given:

$\frac{{\left(x-3\right)}^2}{4}-\frac{y²}{25}=1$

Formula used:

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

Calculation:

Given that $\frac{{\left(x-3\right)}^2}{4}-\frac{y²}{25}=1$.

The equation represents hyperbola.

$\text{Center }=\left(h,k\right)=\left(3,0\right)$

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

$a=2;b=5$

Vertices $(h\pm a,k)=\left(3\pm 2,0\right)=\left(1,0\right),\left(5,0\right)$

$c²=b²+a²=2²+5²=\sqrt{29}$

Foci $\left(h\pm c,k\right)=\left(3\pm \sqrt{29},0\right)$

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

$y=\pm \frac{5}{2}\left(x-3\right)$

The graph of the equation of $\frac{{\left(x-3\right)}^2}{4}-\frac{y²}{25}=1$ is plotted.

Using the given equation $\frac{{\left(x-3\right)}^2}{4}-\frac{y²}{25}=1$, we plot the graph using graphing tool.

The graph represents a hyperbola, with center $\left(3,0\right)$ and the transverse axis is $x\text{-axis}$.