Question

Let $E_1$ and $E_2$ be two ellipse whose centres are at the origin. The major axes of $E_1$ and $E_2$ lie along 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 curves S, $E_1$ and $E_2$ at P, q and R, respectively.
Suppose that $PQ=PR=\frac{2\sqrt{2}}{3}.$ 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. $e_1{e}_2=\frac{\sqrt{7}}{2\sqrt{10}}$
• 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

$e_1{e}_2=\frac{\sqrt{7}}{2\sqrt{10}}$

Here $E_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)E_2:\frac{x^2}{c^2}+\frac{y^2}{d^2}=1,(c<d)andS:x^2+(y-1{)}^2=2$
As tangent to $E_1,E_2$ and $S$ is $x+y=3$

Let the point of contact of tangent be $(x_1,y_1)toS.\therefore x.x_1+y.y_1-(y+y_1)+1=2Orx{x}_1+y{y}_1-y=(1+y_1),issameasx+y=3\Rightarrow \frac{x_1}{1}=\frac{y_1-1}{1}=\frac{1+y_1}{3}i.e.x_1=1and{y}_1=2p=(1,2)PR=PQ=\frac{2\sqrt{2}}{3}thusbyparametricform,\frac{x-1}{-\frac{1}{\sqrt{2}}}=\frac{y-2}{\frac{1}{\sqrt{2}}}=\pm \frac{2\sqrt{2}}{3}\Rightarrow (x=\frac{5}{3},y=\frac{4}{3})And(x=\frac{1}{3},y=\frac{8}{3})$
$\therefore Q=(\frac{5}{3},\frac{4}{3})andR=(\frac{1}{3},\frac{8}{3})$
Now, equation tangent at $Q$ on Ellipse $E_1$ is
$\frac{x.5}{a^2.3}+\frac{y.4}{b^2.3}=1$
On comparing with $x+y=3$, we get
$a^2=5$ and $b^2=4$
$\therefore e_1^2=1-\frac{b^2}{a^2}=1-\frac{4}{5}=\frac{1}{5}$
Also, equation of tangent at R on ellipse $E_2$ is
$\frac{x.1}{a^2.3}+\frac{y.8}{b^2.3}=1$
On comparing with $x+y=3$, we get
$a^2=1,b^2=8$
$\therefore e_2^2=1-\frac{a^2}{b^2}=1-\frac{1}{8}=\frac{7}{8}$
Now, $e_1^2.e_2^2=\frac{7}{40}\Rightarrow e_1{e}_2=\frac{\sqrt{7}}{2\sqrt{10}}$
and $e_1^2+e_2^2=\frac{1}{5}+\frac{7}{8}=\frac{43}{40}$
Also, $|e_1^2-e_2^2|=|\frac{1}{5}-\frac{7}{8}|=\frac{27}{40}$