Question

Finding the Equation of an Ellipse

Find an equation for the ellipse that satisfies the given conditions.

Eccentricity: $\frac{1}{3}$, Foci: $(0,\pm 2)$

Short answer

The equation of ellipse is $\frac{x^2}{25}+\frac{y^2}{50}=1$

Step-by-step solution

Step 1. Given information

Foci is given as $(0,\pm 2)$ and the eccentricity is $\frac{1}{3}$.

Step 2. Concept used

In a horizontal ellipse, the length of the major axis is $2a$ and the length of the minor axis is $2b$.

In a vertical ellipse, the length of the major axis is $2b$ and length of the minor axis is $2a$.

Use the standard form of the equation of the ellipse and the formula is-

$c^2=b^2-a^2$

we will find the equation of an ellipse by substituting the values.

Step 3. Calculation

We know that in $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$where $a>b$

foci are $(0,\pm ae)=(0,\pm 2)$

So we get as $=\pm 2$

Eccentricity is $\frac{1}{3}$

When Foci are $(0,\pm 2)$

$\begin{array}{l}\text{ae}=\pm 2\\ a\left(\frac{1}{3}\right)=\pm 2\\ a=\pm 6\end{array}$

Squaring both sides we get

$a^2=36$

We know that $b^2=a^2\left(1-e^2\right)$

$\begin{array}{l}{\text{b}}^2=36\left(1-{\left(\frac{1}{3}\right)}^2\right)\\ {\text{b}}^2=36\left(1-\frac{1}{9}\right)\\ b^2=36\left(\frac{8}{9}\right)\\ b^2=32\end{array}$

The equation of an ellipse is $\frac{x^2}{32}+\frac{y^2}{36}=1$