Question

Let $E_1$ and $E_2$ be two ellipses whose centers are in origin. The major axes of $E_1$ and $E_2$ lies along the $x-$axis and $y-$axis, respectively. Let $S$ be the circle $x^2+(y-1{)}^2=2.$. The straight lin $x+y=3$ touches the curves $S$, $E_1$ and $E_2$ at $P,Q$ and $R,$ respectively. Supoose the $PQ=PR=\frac{2\sqrt{2}}{3}.$If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$, respectively, then the correct expression(s) is(are)

• A. $e_1^2+e_2^2=\frac{43}{40}$
• B. $e_1{e}_2=\frac{\sqrt{7}}{2\sqrt{1}0}$
• 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 $e_1{e}_2=\frac{\sqrt{7}}{2\sqrt{1}0}$

Let
$E_1:\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}\text{where,}{a}_1>b_1{E}_2:\frac{x^2}{a_2^2}+\frac{y^2}{b_2^2}\text{where,}{a}_2<b_2$
Straight line $x+y=3$ touches the circle $S\to x^2+(y-1{)}^2=2$ at point $P(x,y)$ i.e.
$x^2+(3-x-1{)}^2=2\Rightarrow x=1$
So, $P(x,y)=P(1,2)$
For $Q\text{and}R\frac{x-1}{-\frac{1}{\sqrt{2}}}=\frac{y-2}{\frac{1}{\sqrt{2}}}=\pm \frac{2\sqrt{2}}{3}\therefore Q:(\frac{5}{3},\frac{4}{3})\text{and}R:(\frac{1}{3},\frac{8}{3})$

Let the point of contact is $Q(x,y)$ to ellipse $E_1$ the the equation of tangent is $\frac{x{x}_1}{a_1^2}+\frac{y{y}_1}{b_1^2}=1$ comparing with $x+y=3,$
$Q(x_1,y_1)=(\frac{a_1^2}{3},\frac{b_1^2}{3})=(\frac{5}{3},\frac{4}{3})$
$a_1^2=5,b_1^2=4\Rightarrow e_1=\frac{1}{\sqrt{5}}$
Similarly point of contact is $R(x_2,y_2)$ to ellipse $E_2$
$R(x_2,y_2)=(\frac{a_2^2}{3},\frac{b_2^2}{3})=(\frac{1}{3},\frac{8}{3})$
$a_2^2=1,b_2^2=8\Rightarrow e_2=\frac{\sqrt{7}}{2\sqrt{2}}$
$e_1^2+e_2^2=\frac{1}{5}+\frac{7}{8}=\frac{43}{40}$
$e_1{e}_2=\frac{1}{\sqrt{5}}\times \frac{\sqrt{7}}{2\sqrt{2}}=\frac{\sqrt{7}}{2\sqrt{1}0}$
$|e_1^2-e_2^2|=|\frac{1}{5}-\frac{7}{8}|=\frac{27}{40}$