Question

A point $P$ moves, so that the difference of its distances from $(-ae,0)$ and $(ae,0)$ is $2a$. Then, the locus of $P$ is

• A. $\left(\frac{x^2}{a^2}\right)-\left(\frac{y^2}{a^2\left(e^2-1\right)}\right)=1$
• B. $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{a^2\left(1-e^2\right)}\right)=1$
• C. $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{a^2\left(1+e^2\right)}\right)=1$
• D. $\left(\frac{x^2}{a^2}\right)-\left(\frac{y^2}{a^2\left(1+e^2\right)}\right)=1$

Answer 1

The correct option is A

$\left(\frac{x^2}{a^2}\right)-\left(\frac{y^2}{a^2\left(e^2-1\right)}\right)=1$

Explanation for the correct option:

Step 1. Let $A(-ae,0)$ and $B(ae,0)$ be the given points and Let $P(h,k)$ be a point such that

$PA–PB=2a$

$\Rightarrow$$\sqrt{{(h+ae)}^2+{(k–0)}^2}–\sqrt{{(h–ae)}^2+{(k–0)}^2}=2a$

$\Rightarrow$ $\sqrt{{(h+ae)}^2+k^2}=2a+\sqrt{{(h–ae)}^2+k^2}$

Step 2. By Squaring both sides, we get

$\Rightarrow$$h^2+a^2{e}^2+2aeh+k^2=4a^2+h^2+a^2{e}^2–2aeh+k^2+4a\sqrt{{(h–ae)}^2+k^2}$

$\Rightarrow$ $aeh=2a^2–aeh+2a\sqrt{{(h–ae)}^2+k^2}$

$\Rightarrow$ $eh–a=\sqrt{{(h–ae)}^2+k^2}$ $\left[\because a\ne 0\right]$

Step 3. By Squaring both sides again, we get

$\Rightarrow$$e^2{h}^2+a^2–2aeh=h^2+a^2{e}^2–2aeh+k^2$

$\Rightarrow$ $e^2{h}^2+a^2=h^2+a^2{e}^2+k^2$

$\Rightarrow$ $a^2(e^2–1)=h^2(e^2–1)–k^2$

$\Rightarrow$ $\frac{h^2}{a^2}–\frac{k^2}{a^2(e^2–1)}=1$

$\Rightarrow$$\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}–\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2({\mathrm{e}}^2–1)}=1$

$\therefore$The locus of $P(h,k)$ is $\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{}\mathbf{–}\mathbf{}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}\mathbf{}\mathbf{(}{\mathbf{e}}^{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}\mathbf{1}\mathbf{)}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$

Hence, Option ‘A’ is Correct.