Chapter 8, Problem 41RE

Solution

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

The standard equation of the ellipse with the vertices $\left(\pm 13,0\right)$ is $\frac{x^2}{169}+\frac{y^2}{25}=1$ .

Given data:

Type of conic: Ellipse

Center of the ellipse $=\left(0,0\right)$

Foci of the ellipse $=\left(\pm 12,0\right)$

Vertices of the ellipse $=\left(\pm 13,0\right)$

Formula used:

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

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

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

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

Calculation:

It is given that the foci and the vertices of the ellipse are $\left(\pm 12,0\right)$ and $\left(\pm 13,0\right)$

respectively. Then, it can be written as:

  $c=12,a=13$

Calculate $b$ using the Pythagorean relation.

  $\begin{array}{c}{a}^2=b^2+c^2\\ {13}^2=b^2+{12}^2\\ b^2=\left(13-12\right)\left(13+12\right)\\ b^2=25\\ b=\pm 5\end{array}$

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

  $\begin{array}{c}\frac{x^2}{{\left(13\right)}^2}+\frac{y^2}{{\left(5\right)}^2}=1\\ \frac{x^2}{169}+\frac{y^2}{25}=1\end{array}$

Therefore, the standard equation of the ellipse with the vertices $\left(\pm 13,0\right)$ is $\frac{x^2}{169}+\frac{y^2}{25}=1$ .