Chapter 8, Problem 45RE

Solution

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

The standard equation of the hyperbola with the vertices $\left(0,\pm 5\right)$ is $\frac{y^2}{25}-\frac{x^2}{11}=1$

Given data:

A hyperbola is given with center $\left(0,0\right)$ , foci $\left(0,\pm 6\right)$ and vertices $\left(0,\pm 5\right)$ .

Formula used:

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

  $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

Foci of the ellipse is $\left(0,\pm c\right)$ .

Vertices of the ellipse is $\left(0,\pm a\right)$ .

Calculation:

It is given that the foci and the vertices of the ellipse are $\left(0,\pm 6\right)$ and $\left(0,\pm 5\right)$

respectively. Then, it can be written as:

  $c=6,a=5$

Calculate $b$ using the Pythagorean relation.

  $\begin{array}{c}{a}^2=b^2+c^2\\ 5^2=b^2+6^2\\ b^2=\left(6-5\right)\left(6+5\right)\\ b^2=11\\ b=\pm \sqrt{11}\end{array}$

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

  $\begin{array}{c}\frac{y^2}{5^2}-\frac{x^2}{{\left(\sqrt{11}\right)}^2}=1\\ \frac{y^2}{25}-\frac{x^2}{11}=1\end{array}$

Therefore, the standard equation of the hyperbola with the vertices $\left(0,\pm 5\right)$ is $\frac{y^2}{25}-\frac{x^2}{11}=1$