Question

Find the equation of the ellipse with center origin and axes as the axes of coordinate whose minor axis is equal to the distance between the foci and whose latus rectum is 3.

Answer 1

The equations of the ellipse are $x^2 + 2y^2 = 9$ and $2x^2 + y^2 = 9$.

Answer 2

Let $a$ be the semi-major axis and $b$ be the semi-minor axis. Given $2b = 2ae \implies b=ae$. Given $\frac{2b^2}{a} = 3$. From $b=ae$, $b^2=a^2e^2$. Using $b^2=a^2(1-e^2)$, we get $a^2e^2=a^2(1-e^2) \implies e^2 = 1/2$. From $\frac{2b^2}{a}=3$, $2b^2=3a$. Substitute $b^2=a^2e^2$: $2a^2e^2=3a \implies 2ae^2=3$. Substitute $e^2=1/2$: $2a(1/2)=3 \implies a=3$. Then $2b^2=3a = 3(3)=9 \implies b^2=9/2$. Since $a=3$ and $b^2=9/2$, $a$ is the semi-major axis and $b=\sqrt{9/2}$ is the semi-minor axis. The possible equations are $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ or $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$. $\frac{x^2}{9} + \frac{y^2}{9/2} = 1 \implies x^2 + 2y^2 = 9$. $\frac{x^2}{9/2} + \frac{y^2}{9} = 1 \implies 2x^2 + y^2 = 9$.