Let $E1$ and $E2$ be two ellipses whose centeres are at the origin. The major axes of $E1 and E2$ 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, E1 and E2 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 E1 and E2 respectively, then the correct expression( s) is are

The equation to the ellipses are 

$E_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)\text{ and }{E}_2:\frac{x^2}{c^2}+\frac{y^2}{d^2}=1(c<d)$

The tangent is $x+y=3$

The tangent at  $(\alpha ,\beta )\text{ to }{x}^2+(y-1{)}^2=2$

$x\alpha +(y-1)(\beta -1)=2$

$\text{ i.e. }x\alpha +(\beta -1)y=\beta +1$

Given ,$\alpha =1,\beta =2$

Hence P is (1, 2). 

$x+y=3$ touches $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ at }\left(x_1,y_1\right)$

$\frac{x{x}_1}{a}+\frac{y{y}_1}{b}=1$

Comparing , we get  $Q\left(\frac{a}{3},\frac{b}{3}\right)$

Again it touches  $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ we get $R\left(\frac{c}{3},\frac{d}{3}\right)$

${\left(\frac{a}{3}-1\right)}^2+{\left(\frac{b}{3}-2\right)}^2=\frac{8}{9}$

Also, $a+b=9,$ we get  $(a,b)=(5,4)$

similarly, $(c,d)=(1,8)$ 

$e_1^2=1-\frac{4}{5}=\frac{1}{5};e_2^2=1-\frac{1}{8}=\frac{7}{8}$