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 \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]

Answer 1

Let \[A\left( - ae, 0 \right)\text{ and }B\left( ae, 0 \right)\] be the given points. Let P(h, k) be a point such that

\[PA - PB = 2a\]

\[\therefore \sqrt{\left( h + ae \right)^2 + \left( k - 0 \right)^2} - \sqrt{\left( h - ae \right)^2 + \left( k - 0 \right)^2} = 2a\]
\[ \Rightarrow \sqrt{\left( h + ae \right)^2 + k^2} = 2a + \sqrt{\left( h - ae \right)^2 + k^2}\]
Squaring both sides, we get:
\[h^2 + a^2 e^2 + 2aeh + k^2 = 4 a^2 + h^2 + a^2 e^2 - 2aeh + k^2 + 4a\sqrt{\left( h - ae \right)^2 + k^2}\]
\[ \Rightarrow aeh = 2 a^2 - aeh + 2a\sqrt{\left( h - ae \right)^2 + k^2}\]
\[ \Rightarrow eh - a = \sqrt{\left( h - ae \right)^2 + k^2} \left( \because a \neq 0 \right)\]
Squaring both sides again, we get:
\[e^2 h^2 + a^2 - 2aeh = h^2 + a^2 e^2 - 2aeh + k^2 \]
\[ \Rightarrow e^2 h^2 + a^2 = h^2 + a^2 e^2 + k^2 \]
\[ \Rightarrow a^2 \left( e^2 - 1 \right) = h^2 \left( e^2 - 1 \right) - k^2 \]
\[ \Rightarrow \frac{h^2}{a^2} - \frac{k^2}{a^2 \left( e^2 - 1 \right)} = 1\]
Hence, the locus of (h, k) is

\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\text{ where }b^2 = a^2 \left( e^2 - 1 \right)\]