Question

Let $E_1$ and $E_2$ be the two ellipses centred at origin. The major axis of $E_1$ and $E_2$ lie along the $x-$ axis and $y-$ axis respectively. Let $S$ be the circle $x^2+(y-1{)}^2=2,$ the straight line $x+y=3$ touches the curve $S,E_1$ and $E_2$ at $P,Q$ and $R$ respectively such that $PQ=PR=\frac{2\sqrt{2}}{3}.$ If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2,$ then which of the following is/are correct

• A. $e_1^2+e_2^2=\frac{43}{40}$
• B. $e_1{e}_2=\frac{\sqrt{7}}{2\sqrt{10}}$
• C. $|e_1^2-e_2^2|=\frac{5}{8}$
• D. $e_1{e}_2=\frac{\sqrt{3}}{4}$

Answer 1

Answer: (A), (B)

The correct options are
A $e_1^2+e_2^2=\frac{43}{40}$
B $e_1{e}_2=\frac{\sqrt{7}}{2\sqrt{10}}$

Let $E_1:\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}-1=0(a_1>b_1)$
$E_2:\frac{x^2}{a_2^2}+\frac{y^2}{b_2^2}-1=0(b_2>a_2)$
$S:x^2+(y-1{)}^2=2$
or $x^2+y^2-2y-1=0$
Given common tangent is $x+y=3$
Solving with $S$
$\therefore P(x_1,y_1)\equiv (1,2)$
Now any point on line $PQR$ will be of form $(1\pm r\cos \theta ,2\pm r\sin \theta )$
where $\tan \theta =-1,r=\frac{2\sqrt{2}}{3}$
$\Rightarrow R\equiv (1/3,8/3),Q\equiv (5/3,4/3)$


Now equation of tangent at $R$ on $E_2$ will be
$\frac{x}{3a_2^2}+\frac{8y}{3b_2^2}=1\Rightarrow \frac{x}{a_2^2}+\frac{8y}{b_2^2}=3$
comparing with tangent equation
$a_2^2=1,b_2^2=8\Rightarrow e_2=\sqrt{\frac{7}{8}}$
similarly with ellipse $E_1$
$a_1^2=5,b_1^2=4\Rightarrow e_1=\sqrt{\frac{1}{5}}$