Chapter 9.2, Problem 20E

To determine

To find:The standard equation of the ellipse with center at origin whenvertices are $\left(\pm 8,0\right)$ and passes through the point $\left(5,-3\right)$ .

Answer to Problem 20E

The equation of the ellipse is $\frac{x^2}{64}+\frac{13y^2}{192}=1$ .

Explanation of Solution

Given information:

The vertices of the ellipse is $\left(\pm 8,0\right)$ andellipse passes through the point $\left(5,-3\right)$ .

Calculation:

The major axis of the ellipse is horizontal. So, the general equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .

Substitute $8$ for $a$ in the general equation of ellipse.

  $\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \frac{x^2}{8^2}+\frac{y^2}{b^2}=1\end{array}$

Substitute $5$ for $x$ and $-3$ for $y$ in the equation as ellipse pass through the point $\left(5,-3\right)$ .

  $\begin{array}{c}\frac{5^2}{8^2}+\frac{{\left(-3\right)}^2}{b^2}=1\\ \frac{25}{64}+\frac{9}{b^2}=1\\ \frac{9}{b^2}=1-\frac{25}{64}\\ \frac{9}{b^2}=\frac{39}{64}\end{array}$

Further simplify it.

  $\begin{array}{c}\frac{9}{b^2}=\frac{39}{64}\\ b^2=\frac{64\times 9}{39}\\ b^2=\frac{192}{13}\end{array}$

Substitute the values in the general equation of the ellipse.

  $\begin{array}{c}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \frac{x^2}{64}+\frac{y^2}{\frac{192}{13}}=1\\ \frac{x^2}{64}+\frac{13y^2}{192}=1\end{array}$

Therefore, the equation of the ellipse is $\frac{x^2}{64}+\frac{13y^2}{192}=1$ .