Chapter 9.6, Problem 5WE

To determine

To calculate: To find the equation of the ellipse

Answer to Problem 5WE

Equation of ellipse is $\frac{{\left(x-2\right)}^2}{25}+\frac{{\left(y+3\right)}^2}{41}=1$

Explanation of Solution

Given information: Foci are $\left(-2,-3\right)$ and $\left(6,-3\right)$ and sum of the focal radii is $10$

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:

Foci of ellipse is given as follows:

  $\left(-2,-3\right)$ and $\left(6,-3\right)$

On comparing the above foci with $\left(h,k\pm c\right)$ :

  $\begin{array}{l}k=-3\\ h-c=-2---(1)\\ h+c=6---(2)\end{array}$

Adding (1) and (2)

  $\begin{array}{l}2h=4\\ h=2\end{array}$

Substituting the value in (1)

  $\begin{array}{l}2-c=-2\\ c=4\end{array}$

Also, given the sum of the focal radii is $10$

  $2a=10$

On solving,

  $\begin{array}{l}a=\frac{10}{2}\\ a=5\end{array}$

Also, $b^2$ is calculated as follows:

  $b^2=c^2+a^2$

Substituting the values:

  $\begin{array}{l}{b}^2=4^2+5^2\\ b^2=16+25\\ b^2=41\\ b=\pm \sqrt{41}\end{array}$

Equation of ellipse is given as:

  $\begin{array}{l}\frac{{\left(x-2\right)}^2}{5^2}+\frac{{\left(y-\left(-3\right)\right)}^2}{{\left(\sqrt{41}\right)}^2}=1\\ \frac{{\left(x-2\right)}^2}{25}+\frac{{\left(y+3\right)}^2}{41}=1\end{array}$

Conclusion:

Hence, equation of ellipse is $\frac{{\left(x-2\right)}^2}{25}+\frac{{\left(y+3\right)}^2}{41}=1$