Show that $PQ$ has length $\frac{2b^2}{a}$

Question

For $P(a\cos(\theta),b\sin(\theta))$ and $Q(a\cos(-\theta), b\sin(-\theta))$, which are extremities of the Latus rectum $x = ae$ of ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$, show that $PQ$ has length $\frac{2b^2}{a}$.

Answer 1

Put $x=ae=\sqrt{a^2-b^2}$ into $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$,

$$\frac{a^2-b^2}{a^2}+\frac{y^2}{b^2}=1$$

$$y^2=\frac{b^4}{a^2}$$

$$y=\pm \frac{b^2}{a}$$

$$PQ=|y_1-y_2|=\frac{2b^2}{a}$$

The "extremities" mean the end points (of the latus rectum).