JEE Advanced 2015 Paper 2Question 1
Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis, respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S$, $E_1$ and $E_2$ at $P$, $Q$ and $R$, respectively. Suppose that $PQ=PR=\frac{2\sqrt2}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$, respectively, then the correct expression(s) is(are)
Ellipse
Solution
$P=(1,2)$. $Q,R=(1\pm\frac23,2\mp\frac23)$. Tangent to $\frac{x^2}{A}+\frac{y^2}{B}=1$ at $(x_0,y_0)$ gives $A=3x_0$, $B=3y_0$: $E_1:\frac{x^2}5+\frac{y^2}4=1$, $E_2:x^2+\frac{y^2}8=1$. $e_1^2=\frac15$, $e_2^2=\frac78$; $e_1^2+e_2^2=\frac{43}{40}$, $e_1e_2=\sqrt{7/40}=\frac{\sqrt7}{2\sqrt{10}}$.