Chapter 10, Problem 38E

Solution

To determine : The equation of a hyperbola whose foci are $\left(0,\pm 25\right)$ and vertices are $\left(0,\pm 7\right)$ .

The required equation is $\frac{y^2}{49}-\frac{x^2}{576}=1$ .

Given information :

The foci of a hyperbola are $\left(0,\pm 25\right)$ and vertices are $\left(0,\pm 7\right)$ .

Formula used:

If the vertices are $\left(0,\pm a\right)$ then the equation of the hyperbola in standard form is

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

Here, $c^2=a^2+b^2$ and $\left(0,\pm c\right)$ are foci.

Asymptotes are $y=\pm \frac{a}{b}x$ .

Explanation :

Here, the foci is on vertical axis, so the equation of hyperbola will be in the form of

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

The vertices are $\left(0,\pm 7\right)$ and the foci are $\left(0,\pm 25\right)$ .

So, $a=7$ and $c=25$ .

  $\begin{array}{c}{\left(25\right)}^2={\left(7\right)}^2+b^2\\ b=24\end{array}$

So, the equation of the hyperbola will be

  $\begin{array}{c}\frac{y^2}{{\left(7\right)}^2}-\frac{x^2}{{\left(24\right)}^2}=1\\ \frac{y^2}{49}-\frac{x^2}{576}=1\end{array}$

Therefore, the equation of parabola is $\frac{y^2}{49}-\frac{x^2}{576}=1$ .