Question

Finding the Equation of an Ellipse

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

End points of major axis: $(\pm 10,0)$, distance between foci is: $6$

Short answer

The equation of ellipse is $\frac{x^2}{100}+\frac{y^2}{91}=1$

Step-by-step solution

Step 1. Given information

End points of major axis $(\pm 10,0)$and the distance between foci is $6$.

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 are given end points of major axis as: $(\pm 10,0)$

$\Rightarrow$It is a horizontal ellipse.

$\begin{array}{l}\Rightarrow a=10\\ \Rightarrow a^2=100\end{array}$

Also, we are given that the distance between foci is 6

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

Using formula-

$\begin{array}{l}{c}^2=a^2-b^2\\ \Rightarrow 3^2={10}^2-b^2\\ \Rightarrow 9=100-b^2\\ \Rightarrow b^2=100-9\\ \Rightarrow b^2=91\end{array}$

The equation of an ellipse is $\frac{x^2}{100}+\frac{y^2}{91}=1$