Question

Finding the Equation of an Ellipse

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

Foci: $F(\pm \sqrt{15},0)$, vertices: $(\pm 6,0)$

Short answer

the equation of ellipse is $\frac{x^2}{36}+\frac{y^2}{21}=1$

Step-by-step solution

Step 1. Given information

Foci $F(\pm \sqrt{15},0)$ and vertices $(\pm 6,0)$ are given

Step 2. Concept used

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 are given Foci: $F(\pm \sqrt{15},0)$and vertices: $(\pm 6,0)$

$\begin{array}{l}\Rightarrow a=6\\ \Rightarrow c=\sqrt{15}\end{array}$

Thus, the major axis is horizntal.

We will find $b^2$from $c^2=b^2-a^2$

Substituting $a=6$and

$c=\sqrt{15}$and simplifying

$\Rightarrow {(}^{\sqrt{15})2}=6^2-b^2$

$\begin{array}{l}\Rightarrow 15=36-b^2\\ \Rightarrow b^2=36-15\\ \Rightarrow b^2=21\end{array}$

The equation of an ellipse is $\frac{x^2}{36}+\frac{y^2}{21}=1$