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, $SP+SP=6$.
If difference of eccentric angles of points $P$ and $Q$ is 60 , then locus of $A$ is -

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

Answer 1

Answer: (C)

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

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