Chapter 9.4, Problem 22WE

To determine

To calculate: To write the equation of ellipse having the given points as foci and the given number as the sum of focal radii

Answer to Problem 22WE

Equation of ellipse is $\frac{x^2}{225}+\frac{y^2}{144}=1$

Explanation of Solution

Given information: Foci are $\left(-9,0\right),\left(9,0\right)$ and sum of focal radii $30$

Formula Used:

Standard form of the equation of ellipse with center $\left(h,k\right)$ and major axis parallel to x-axis is:

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

Where, $a=$ x-intercept, $b=$ y-intercept, $c^2=a^2-b^2$

When, $a>b$ , foci lies on x-axis and coordinates of foci are $\left(h\pm c,k\right)$

When $b>a$ , foci lies on y-axis and coordinates of foci are $\left(h,k\pm c\right)$

Calculation:

Given the Foci as follows $\left(-9,0\right),\left(9,0\right)$

Thus, $c=\pm 9$

Also, given the sum of focal radii $30$

  $\begin{array}{l}2a=30\\ a=\frac{30}{2}\\ a=15\end{array}$

Now, $b$ is calculated using the below formula:

  $c^2=a^2-b^2$

Substituting the values:

  $\begin{array}{l}{b}^2=a^2-c^2\\ b^2={15}^2-9^2\\ b^2=225-81\\ b^2=144\\ b=\pm 12\end{array}$

Equation of ellipse is given as follows:

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

Substituting the values:

  $\begin{array}{l}\frac{{\left(x-0\right)}^2}{{15}^2}+\frac{{\left(y-0\right)}^2}{{12}^2}=1\\ \frac{x^2}{225}+\frac{y^2}{144}=1\end{array}$

Conclusion:

Hence, equation of ellipse is $\frac{x^2}{225}+\frac{y^2}{144}=1$