Question

Finding the Equation of an Ellipse

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

Eccentricity: $0.75$, Foci: $(\pm 1.5,0)$

Short answer

The equation of ellipse is $\frac{x^2}{4}+\frac{y^2}{1.75}=1$

Step-by-step solution

Step 1. Given information

Foci is given as $(\pm 1.5,0)$ and the eccentricity is $0.75$.

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 $(\pm 1.5,0)$

So we get$ae=\pm 2$

Eccentricity is $0.75$

When Foci are $(\pm 1.5,0)$

$\begin{array}{l}ae=\pm 1.5\\ a\left(0.75\right)=\pm 1.5\\ a=\pm 2\end{array}$

Squaring both sides we get

$a^2=4$

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

$\begin{array}{l}{b}^2=4\left(1-{\left(0.75\right)}^2\right)\\ {\text{b}}^2=1.75\end{array}$

The equation of an ellipse is $\frac{x^2}{4}+\frac{y^2}{1.75}=1$