Question

Let $S,S^{\prime }$ be the foci of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ whose eccentricity is '$e$'. $P$ is a variable point on the ellipse. Consider the locus of the incentre of the $△PS{S}^{\prime }$

Maximum area of rectangle inscribed in the locus is

• A. $\frac{2ab{e}^2}{1+e}$
• B. $\frac{2abe}{1-e}$
• C. $\frac{abe}{1+e}$
• D. None of these

Answer 1

The correct option is A $\frac{2ab{e}^2}{1+e}$


Let the coordinates of $P$ be
$(a\cos \theta ,b\sin \theta )$
Here,
$SP=\text{Focal distance of point}P$
So,
$SP=a-ae\cos \theta$
$S^{\prime }P=a+ae\cos \theta$
$S{S}^{\prime }=2ae$

If $(h,k)$ are the coordinates of the incenter of $\Delta PS{S}^{{}^{\prime }}$, then
$h=\frac{2ae\left(a\cos \theta \right)+a(1-e\cos \theta )(-ae)+a(1+e\cos \theta )ae}{2ae+a(1-e\cos \theta )+a(1+e\cos \theta )}=ae\cos \theta \cdots \left(i\right)$
and
$k=\frac{2ae\left(b\sin \theta \right)+a(1-e\cos \theta )\times 0+a(1+e\cos \theta )\times 0}{2ae+a(1-e\cos \theta )+a(1+e\cos \theta )}=\frac{eb\sin \theta }{(e+1)}\cdots \left(ii\right)$
From (i) and (ii),
$\frac{h}{ae}=\cos \theta \frac{k(e+1)}{eb}=\sin \theta$
Eliminating $\theta$, we get
$\frac{x^2}{a^2{e}^2}+\frac{y^2}{{\left(\frac{be}{(e+1)}\right)}^2}=1$
Which clearly represents an ellipse.

Maximum area of rectangle is
$2\left(ae\right)\left(\frac{be}{e+1}\right)=\frac{2ab{e}^2}{e+1}$