Question

The locus of mid-points of focal chords of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity $e$ is

• A. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{ex}{a}$
• B. $\frac{x^2}{a^2}-\frac{y^2}{b^2}=\frac{ex}{a}$
• C. $x^2+y^2=a^2+b^2$
• D. None of the above

Answer 1

The correct option is A $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{ex}{a}$


Given $S:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let $P(h,k)$ be the mid-point of the focal chord. Then its equation is $T=S_1$
$\Rightarrow \frac{hx}{a^2}+\frac{ky}{b^2}-1=\frac{h^2}{a^2}+\frac{k^2}{b^2}-1$
$\Rightarrow \frac{hx}{a^2}+\frac{ky}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$

Since, it is a focal chord, so it passes through focus, either $(ae,0)$ or $(–ae,0).$
If it passes through $(ae,0),$ then
$\Rightarrow \frac{h\left(ae\right)}{a^2}+0=\frac{h^2}{a^2}+\frac{k^2}{b^2}$
Locus of point $P$ is
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{ex}{a}$

If it passes through $(-ae,0),$ then
$\Rightarrow \frac{h(-ae)}{a^2}+0=\frac{h^2}{a^2}+\frac{k^2}{b^2}$
Locus of point $P$ is
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=-\frac{ex}{a}$