Chapter 10.4, Problem 46E

To determine

To find:the standard form of the equation of the hyperbola.

Answer to Problem 46E

Thestandard form of the equation of the hyperbola is $\frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y-3\right)}^2}{9}=1$ .

Explanation of Solution

Given information:

The given vertices are $\left(3,0\right),\left(3,4\right)$

The given asymptotes are $y=\frac{2}{3}x,y=4-\frac{2}{3}x$ .

Calculation:

The vertices are $\left(3,0\right),\left(3,4\right)$ .

The asymptotes are $y=\frac{2}{3}x,y=4-\frac{2}{3}x$ .

The transverse axis of the hyperbola is horizontal.

Calculate the standard equation of the hyperbola.

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

The centeris at $\left(h,k\right)$ , vertices are at $\left(h,k-a\right)and\left(h,k+a\right)$ and foci at $\left(h,k-c\right),and\left(h,k+c\right)$ .

The asymptotes are at $y=\pm \frac{a}{b}\left(x-h\right)+k$ .

In the hyperbola the vertices are at $\left(h,k-a\right)=\left(3,0\right)$ and $\left(h,k+a\right)=\left(3,4\right)$ .

Then $k=2,h=3,a=2$ .

The given asymptotes are $y=\frac{2}{3}x,y=4-\frac{2}{3}x$ .

  $\begin{array}{l}\frac{a}{b}=\frac{2}{3}\\ b=\frac{3a}{2}\\ b=3\end{array}$ .

The standard form of the equation.

  $\begin{array}{c}\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1\\ \frac{{\left(y-2\right)}^2}{{\left(2\right)}^2}-\frac{{\left(x-3\right)}^2}{{\left(3\right)}^2}=1\\ \frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y-3\right)}^2}{9}=1\end{array}$

Therefore, thestandard form of the equation of the hyperbola is $\frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y-3\right)}^2}{9}=1$ .