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 $PQ = PR = \frac{2\sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$, respectively, then the correct expression(s) is (are)

Let E₁ and E₂ be two ellipses whose centers 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 x² + (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 = 2√2/3. If e₁ and e₂ are the eccentricities of E₁ and E₂, respectively, then the correct expression(s) is (are)

$E_1, E_2: \text{ ellipses centred at the origin}$

$E_1 \text{ major axis along } x\text{-axis}; \ E_2 \text{ along } y\text{-axis}$

$S: x^2 + (y-1)^2 = 2$

$x+y=3 \text{ touches } S, E_1, E_2 \text{ at } P, Q, R$

$PQ = PR = \frac{2\sqrt{2}}{3}$

$\text{To find: correct option(s) for } e_1, e_2$

$C(0,1),\quad r = \sqrt2$ $L:\ x+y=3$ $1.\ \text{tangency } L, S \Rightarrow P$ $2.\ Q, R \text{ on } L,\ PQ=PR \Rightarrow \text{coords}$ $3.\ \text{tangency } L, E_1, E_2 \Rightarrow e_1, e_2$ $d(C,L) = \dfrac{|0+1-3|}{\sqrt2} = \sqrt2 = r$ $\Rightarrow L \text{ is tangent to } S$ $P = C + \dfrac{2}{2}(1,1) = (0,1)+(1,1) = (1,2)$ $CP \perp L$ $PQ = PR = \frac{2\sqrt{2}}{3}$ $\hat u = \frac{(1,-1)}{\sqrt2}$ $d\,\hat u = \frac{2\sqrt2}{3}\cdot\frac{(1,-1)}{\sqrt2} = \left(\tfrac23,-\tfrac23\right)$ $Q,R = (1,2)\pm\left(\tfrac23,-\tfrac23\right) = \left(\tfrac53,\tfrac43\right),\ \left(\tfrac13,\tfrac83\right)$ $\text{ellipse } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $\text{tangent at } (x_1,y_1):\ \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$ $x_1=\frac{a^2}{3},\ y_1=\frac{b^2}{3} \Rightarrow \text{contact }\left(\frac{a^2}{3},\frac{b^2}{3}\right)$ $\frac{a^2}{9}+\frac{b^2}{9}=1 \Rightarrow a^2+b^2=9$ $E_1:\ a_1^2+b_1^2=9$ $Q=\left(\frac{a_1^2}{3},\frac{b_1^2}{3}\right),\quad a_1>b_1$ $E_2:\ \frac{x^2}{b_2^2}+\frac{y^2}{a_2^2}=1,\quad a_2^2+b_2^2=9$ $R=\left(\frac{b_2^2}{3},\frac{a_2^2}{3}\right),\quad a_2>b_2$ $Q=\left(\tfrac53,\tfrac43\right) \Rightarrow a_1^2=5,\ b_1^2=4$ $R=\left(\tfrac13,\tfrac83\right) \Rightarrow b_2^2=1,\ a_2^2=8$ $e_1^2 = 1-\frac{b_1^2}{a_1^2} = 1-\frac{4}{5} = \frac15$ $e_2^2 = 1-\frac{b_2^2}{a_2^2} = 1-\frac{1}{8} = \frac78$ $e_1^2+e_2^2 = \frac15+\frac78 = \frac{8+35}{40} = \frac{43}{40} \ \checkmark$ $e_1 e_2 = \sqrt{\frac15\cdot\frac78} = \sqrt{\frac{7}{40}} = \frac{\sqrt7}{2\sqrt{10}} \ \checkmark$ $\left|e_1^2-e_2^2\right| = \left|\frac15-\frac78\right| = \frac{27}{40} \neq \frac58 \ \times$ $e_1 e_2 = \frac{\sqrt7}{2\sqrt{10}} \neq \frac{\sqrt3}{4} \ \times$ $\boxed{\text{Correct options: (1) and (2)}}$