Chapter 9, Problem 6CT

To determine

To find: Equation of Hyperbola.

Answer to Problem 6CT

  $\frac{y^2}{9}-\frac{x^2}{4}=1$

Explanation of Solution

Given information:

Vertices = $\left(0,3\right)and\left(0,-3\right)$

  $y=\pm \frac{3}{2}x$

  $y=\pm \frac{3}{2}x$ are asymptotes

Slope of Asymptote = $\pm \frac{3}{2}$

The line $y=\frac{3}{2}x$ passes through origin.

The line $y=-\frac{3}{2}x$ also passes through Origin

Therefore, the intersection point of these two lines is Origin that is $\left(0,0\right)$ .

The intersection of Asymptotes gives center of the Hyperbola.

Center or Hyperbola = $\left(0,0\right)$ = $\left(h,k\right)$

The distance between Center and Vertices is Length of Semi Transverse axis

Length of Semi Transverse axis = $\sqrt{{\left(0-0\right)}^2+{\left(0-3\right)}^2}$

  $\begin{array}{l}\Rightarrow a=\sqrt{{\left(-3\right)}^2}\\ \Rightarrow a=3\end{array}$

The Center origin and the vertices $\left(0,\pm 3\right)$ lies on Y-axis. Therefore, the transverse axis of Hyperbola is Vertical.

For a Hyperbola with Vertical Transverse axis,

The slope of two asymptotes are $\frac{a}{b}$ and $\frac{-a}{b}$

Therefore

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

The standard form of the Hyperbola with center $\left(h,k\right)$ and with Vertical transverse axis and with length of semi-major axis $a$ and length of semi-minor axis $b$ is

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

Therefore, the required equation of Hyperbola is,

  $\Rightarrow \frac{{\left(y-0\right)}^2}{3^2}-\frac{{\left(x-0\right)}^2}{2^2}=1$

  $\Rightarrow \frac{y^2}{9}-\frac{x^2}{4}=1$