Question

A point moves so that the difference of its distances from (ae, 0) and (−ae, 0) is 2a. Prove that the equation to its locus is
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, where b² = a² (e² − 1).

Answer 1

Let $A\left(-ae,0\right)\mathrm{and}B\left(ae,0\right)$ be the given points. Let P(h, k) be a point such that $PA-PB=2a$.

$$\begin{gathered} \therefore \sqrt{{\left(h+ae\right)}^2+{\left(k-0\right)}^2}-\sqrt{{\left(h-ae\right)}^2+{\left(k-0\right)}^2}=2a \\ \Rightarrow \sqrt{{\left(h+ae\right)}^2+k^2}=2a+\sqrt{{\left(h-ae\right)}^2+k^2} \end{gathered}$$

Squaring both sides, we get:

$$\begin{gathered} h^2+a^2{e}^2+2aeh+k^2=4a^2+h^2+a^2{e}^2-2aeh+k^2+4a\sqrt{{\left(h-ae\right)}^2+k^2} \\ \Rightarrow aeh=2a^2-aeh+2a\sqrt{{\left(h-ae\right)}^2+k^2} \\ \Rightarrow eh-a=\sqrt{{\left(h-ae\right)}^2+k^2}\left(\because a\ne 0\right) \end{gathered}$$

Squaring both sides again, we get:

$$\begin{gathered} e^2{h}^2+a^2-2aeh=h^2+a^2{e}^2-2aeh+k^2 \\ \Rightarrow e^2{h}^2+a^2=h^2+a^2{e}^2+k^2 \\ \Rightarrow a^2\left(e^2-1\right)=h^2\left(e^2-1\right)-k^2 \\ \Rightarrow \frac{h^2}{a^2}-\frac{k^2}{a^2\left(e^2-1\right)}=1 \end{gathered}$$

Hence, the locus of (h, k) is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\mathrm{where}{b}^2=a^2\left(e^2-1\right)$.