Chapter 9.2, Problem 19E

To determine

To find:The standard equation of the ellipse with center at origin when vertices are $\left(0,\pm 4\right)$ and passes through the point $\left(3,1\right)$ .

Answer to Problem 19E

The equation of the ellipse is $\frac{5x^2}{48}+\frac{y^2}{16}=1$ .

Explanation of Solution

Given information:

The vertices of the ellipse is $\left(0,\pm 4\right)$ andellipse passes through the point $\left(3,1\right)$ .

Calculation:

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

Substitute $4$ for $a$ in the general equation of ellipse.

  $\begin{array}{l}\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\\ \frac{x^2}{b^2}+\frac{y^2}{4^2}=1\end{array}$

Substitute $3$ for $x$ and $1$ for $y$ in the equation as ellipse pass through the point $\left(3,1\right)$ .

  $\begin{array}{c}\frac{3^2}{b^2}+\frac{1^2}{4^2}=1\\ \frac{9}{b^2}+\frac{1}{16}=1\\ \frac{9}{b^2}=1-\frac{1}{16}\\ \frac{9}{b^2}=\frac{15}{16}\end{array}$

Further simplify it.

  $\begin{array}{c}\frac{9}{b^2}=\frac{15}{16}\\ b^2=\frac{16\times 9}{15}\\ b^2=\frac{48}{5}\end{array}$

Substitute the values in the general equation of the ellipse.

  $\begin{array}{c}\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\\ \frac{x^2}{\frac{48}{5}}+\frac{y^2}{16}=1\\ \frac{5x^2}{48}+\frac{y^2}{16}=1\end{array}$

Therefore, the equation of the ellipse is $\frac{5x^2}{48}+\frac{y^2}{16}=1$ .