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 $△PS{S}^{\prime }$

The eccentricity of locus of $P$ is

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

Answer 1

The correct option is B $\sqrt{\frac{2e}{1+e}}$


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.

Let $e_1$ be its eccentricity.
Then
$\frac{b^2{e}^2}{(e+1{)}^2}=a^2{e}^2(1-e_1^2)$
$\Rightarrow e_1^2=1-\frac{b^2}{a^2(e+1{)}^2}$
$\Rightarrow e_1^2=1-\frac{1-e^2}{(e+1{)}^2}=1-\frac{1-e}{1+e}$
$\Rightarrow e_1^2=\frac{2e}{e+1}$
$\Rightarrow e_1=\sqrt{\frac{2e}{e+1}}$