Question

If a variable point $P$ on an ellipse with eccentricity $e$ is joined to it's focii $S,S^{\prime }$.Then locus of incentre of $△PS{S}^{\prime }$ is another ellipse whose eccentricity is

• A. $\sqrt{\frac{2e}{1+e}}$
• B. $\sqrt{\frac{2e}{1-e}}$
• C. $\sqrt{\frac{e}{1-e}}$
• D. $\sqrt{\frac{e}{1+e}}$

Answer 1

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

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ be an ellipse

Let $P(a\cos \theta ,b\sin \theta )$ be any point on ellipse ,
$SP=a-ex=a-ae\cos \theta$
$S^{\prime }P=a+ex=a+ae\cos \theta$

Also, $S{S}^{\prime }=2ae$
If $(h,k)$ is incentre of $△PS{S}^{\prime }$


Then $h=\frac{\left(2ae\right)a\cos \theta +a(1-e\cos \theta )(-ae)+a(1+e\cos \theta )\left(ae\right)}{2ae+a(1-e\cos \theta )+a(1+e\cos \theta )}$
$h=ae\cos \theta$

Similarly, $k=\frac{2ae\left(b\sin \theta \right)+0+0}{2ae+2a}$
$k=\frac{eb\sin \theta }{e+1}$
$\Rightarrow \cos \theta =\frac{h}{ae},\sin \theta =\frac{k(e+1)}{eb}$

$\therefore$ Required locus is
$\frac{h^2}{a^2{e}^2}+\frac{k^2}{{\left(\frac{be}{e+1}\right)}^2}=1$

Above locus clearly represents an ellipse whose eccentricity is $e_1$

$\therefore \frac{b^2{e}^2}{(1+e{)}^2}=a^2{e}^2(1-e_1^2)$
$\Rightarrow e_1^2=1-\frac{b^2}{a^2(1+e{)}^2}$
$=1-\frac{(1-e^2)}{(1+e{)}^2}$
$\Rightarrow e_1=\sqrt{\frac{2e}{1+e}}$