Chapter 9.4, Problem 21WE

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 21WE

Equation of ellipse is $\frac{x^2}{128}+\frac{y^2}{144}=1$

Explanation of Solution

Given information: Foci are $\left(0,4\right),\left(0,-4\right)$ and sum of focal radii $24$

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 $\left(0,4\right),\left(0,-4\right)$

Thus, $c=\pm 5$

Also, given the sum of focal radii $24$

  $\begin{array}{l}2b=24\\ b=\frac{24}{2}\\ b=12\end{array}$

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

  $c^2=a^2-b^2$

Substituting the values:

  $\begin{array}{l}{a}^2=b^2-c^2\\ a^2={12}^2-4^2\\ a^2=144-14\\ a^2=128\\ a=\pm 8\sqrt{2}\end{array}$

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:

  $\begin{array}{l}\frac{{\left(x-0\right)}^2}{{\left(\pm 8\sqrt{2}\right)}^2}+\frac{{\left(y-0\right)}^2}{{\left(12\right)}^2}=1\\ \frac{x^2}{128}+\frac{y^2}{144}=1\end{array}$

Conclusion:

Hence, equation of ellipse is $\frac{x^2}{128}+\frac{y^2}{144}=1$