Chapter 9.6, Problem 1WE

To determine

To calculate: To find the equation of the ellipse

Answer to Problem 1WE

Equation of ellipse is $\frac{{\left(x-6\right)}^2}{25}+\frac{{\left(y-3\right)}^2}{34}=1$

Explanation of Solution

Given information: Foci $\left(6,0\right)$ and $\left(6,6\right)$ 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:

  $\left(6,0\right)$ and $\left(6,6\right)$

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

  $\begin{array}{l}h=6\\ k-c=0---(1)\\ k+c=6---(2)\end{array}$

Adding (1) and (2)

  $\begin{array}{l}2k=6\\ k=3\end{array}$

Substituting the value in (1)

  $\begin{array}{l}3-c=0\\ c=3\end{array}$

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:

  $\begin{array}{l}{b}^2=3^2+5^2\\ b^2=9+25\\ b^2=34\\ b=\pm \sqrt{34}\end{array}$

Equation of ellipse is given as:

  $\begin{array}{l}\frac{{\left(x-6\right)}^2}{5^2}+\frac{{\left(y-3\right)}^2}{{\sqrt{34}}^2}=1\\ \frac{{\left(x-6\right)}^2}{25}+\frac{{\left(y-3\right)}^2}{34}=1\end{array}$

Conclusion:

Hence, equation of ellipse is $\frac{{\left(x-6\right)}^2}{25}+\frac{{\left(y-3\right)}^2}{34}=1$