Chapter 8, Problem 48RE

Solution

To find: the equation for the conic in the standard form.

The standard equation of the hyperbola with center $\left(-5,0\right)$ and focal axis $x=-5$ is $\frac{y^2}{4}-\frac{{\left(x+5\right)}^2}{5}=1$ .

Given data:

A hyperbola is given with center $\left(-5,0\right)$ , a vertex $\left(-5,2\right)$ and focus $\left(-5,3\right)$ .

Formula used:

The standard equation of the hyperbola with center $\left(h,k\right)$ and focal axis $x=h$ is given as:

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

Vertices and foci are $\left(h,k\pm a\right)$ and $\left(h,k\pm c\right)$ .

Calculation:

It is given that the a vertex, focus and center of the hyperbola is $\left(-5,2\right)$ , $\left(-5,3\right)$ and $\left(-5,0\right)$ respectively.

  $h=-5,k=0$

  $\begin{array}{c}k\pm a=2\\ \pm a=2\\ k\pm c=3\\ \pm c=3\end{array}$

Calculate $b$ using Pythagoreon relation.

  $\begin{array}{c}{c}^2=a^2+b^2\\ 3^2=2^2+b^2\\ b^2=\left(3-2\right)\left(3+2\right)\\ b=\pm \sqrt{5}\end{array}$

The standard equation of the hyperbola with center $\left(-5,0\right)$ and focal axis $x=-5$ is given as:

  $\begin{array}{l}\frac{{\left(y-0\right)}^2}{2^2}-\frac{{\left(x-\left(-5\right)\right)}^2}{{\left(\sqrt{5}\right)}^2}=1\\ \frac{y^2}{4}-\frac{{\left(x+5\right)}^2}{5}=1\end{array}$

Therefore, the standard equation of the hyperbola with center $\left(-5,0\right)$ and focal axis $x=-5$ is $\frac{y^2}{4}-\frac{{\left(x+5\right)}^2}{5}=1$ .