Ellipse Equation and Eccentricity

Problem

An ellipse has foci at $(-3,0)$ and $(3,0)$ and passes through $(5,0)$; find its standard form equation, its eccentricity, and the length of its minor axis.

Step 1: Read off $c$ and $a$ from the foci and vertex

With foci on the $x$-axis, the ellipse has center at the origin and standard form

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad c^2=a^2-b^2.$$

The foci are at $(\pm c,0)$, so $c=3$. Since $(5,0)$ lies on the ellipse, it is a vertex, so $a=5$.

Step 2: Find $b$ and write the equation

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

$$9=25-b^2$$

so

$$b^2=16.$$

Therefore the ellipse equation is

$$\frac{x^2}{25}+\frac{y^2}{16}=1.$$

Step 3: Compute eccentricity and minor axis length

The eccentricity is

$$e=\frac{c}{a}=\frac{3}{5}.$$

The minor axis length is

$$2b=2(4)=8.$$

Answer

The ellipse is $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$, its eccentricity is $\dfrac{3}{5}$, and its minor axis length is $8$.