Question

An ellipse whose distance between foci $S$ and $S^{\prime}$ is 4 units is inscribed in the triangle $ABC$ touching the sides $AB , AC$ and $BC$ at $P , Q$ and $R$. If centre of ellipse is at origin and major axis along $x$-axis, $SP + S P =6$.
If difference of eccentric angles of points $P$ and $Q$ is 60 , then locus of $A$ is -

• A. $16 x^2+9 y^2=144$
• B. $16 x^2+45 y^2=576$
• C. $5 x^2+9 y^2=60$
• D. $5 x^2+9 y^2=15$

Answer 1

Answer: (C)

$2 a e=4$
$2 a =6$
$e =2 / 3$
$b ^2= a ^2\left(1-e^2\right)$
$=9\left(1-\frac{4}{9}\right)$
$=5$

Chord of contact of $A ( h , k )$ is
$\frac{h x}{9}+\frac{k y}{5}=1$ ....(1)
$\frac{x}{3} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{y}{\sqrt{5}} \cdot \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)$....(2)
Comparing (1) & (2)
$\frac{ h }{3 \cos \left(\frac{\alpha+\beta}{2}\right)}=\frac{ k }{\sqrt{5} \sin \left(\frac{\alpha+\beta}{2}\right)}=\frac{1}{\cos \left(\frac{\alpha-\beta}{2}\right)}$
$\frac{ h }{3 \cos \left(\frac{\alpha+\beta}{2}\right)}=\frac{ k }{\sqrt{5} \sin \left(\frac{\alpha+\beta}{2}\right)}=\frac{2}{\sqrt{3}}$
$\cos \left(\frac{\alpha+\beta}{2}\right)=\frac{h}{2 \sqrt{3}} ; \sin \left(\frac{\alpha+\beta}{2}\right)=\frac{\sqrt{3} k }{2 \sqrt{5}}$
$\Rightarrow \frac{x^2}{12}+\frac{3 y^2}{20}=1 $
$ \Rightarrow 5 x^2+9 y^2=60$