Let C be the circle $\mathrm{x}^2+(\mathrm{y}-1)^2=2, \mathrm{E}_1$ and $\mathrm{E}_2$ be two ellipses whose…

Let C be the circle $\mathrm{x}^2+(\mathrm{y}-1)^2=2, \mathrm{E}_1$ and $\mathrm{E}_2$ be two ellipses whose centres lie at the origin and major axes lie on x -axis and y -axis respectively. Let the straight line $x+y=3$ touch the curves $C$, $E_1$ and $E_2$ at $P\left(x_1, y_1\right), Q\left(x_2, y_2\right)$ and $R\left(x_3, y_3\right)$ respectively. Given that $P$ is the mid-point of the line segment $Q R$ and $P Q=\frac{2 \sqrt{2}}{3}$, the value of $9\left(x_1 y_1+x_2 y_2+x_3 y_3\right)$ is equal to ______ .

Solution

Let $\mathrm{E}_1: \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1,(\mathrm{a} \gt \mathrm{b})$
$\mathrm{E}_2: \frac{\mathrm{x}^2}{\mathrm{c}^2}+\frac{\mathrm{y}^2}{\mathrm{~d}^2}=1,(\mathrm{c} \lt \mathrm{d})$
$C: x^2+(y-1)^2=2$
Equation of tangent at $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$
$\mathrm{xx}_1+\mathrm{y}\left(\mathrm{y}_1-1\right)=\left(\mathrm{y}_1+1\right)$
comparing with $\mathrm{x}+\mathrm{y}=3$ we get $\mathrm{P}(1,2)$
$\because$ Now parametric equation of $x+y=3$
$\frac{(\mathrm{x}-1)}{\left(\frac{-1}{\sqrt{2}}\right)}=\frac{(\mathrm{y}-2)}{\left(\frac{1}{\sqrt{2}}\right)}= \pm \frac{2 \sqrt{2}}{3} \quad\left(\because \mathrm{PQ}=\frac{2 \sqrt{2}}{3}\right)$
On solving we get $\mathrm{Q}\left(\frac{5}{3}, \frac{4}{3}\right), \mathrm{R}\left(\frac{1}{3}, \frac{8}{3}\right)$
So, $9\left(\mathrm{x}_1 \mathrm{y}_1+\mathrm{x}_2 \mathrm{y}_2+\mathrm{x}_3 \mathrm{y}_3\right)$
$\begin{aligned}
& 9\left(2+\frac{5}{3} \times \frac{4}{3}+\frac{1}{3} \times \frac{8}{3}\right) \\ & \Rightarrow 46
\end{aligned}$

Asked in: JEE Main 2025 (04 Apr Shift 1)

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