Let C be the circle x^ 2 +(y-1)^ 2 =2, E_ 1 and E_ 2 be two ellip… - Vidyadip

Question

Let $C$ be the circle $x^{2}+(y-1)^{2}=2, E_{1}$ and $E_{2}$ be two ellipses whose centres lie at the origin and major axes lie on $x$-axis and $y$-axis respectively. Let the straight line $\mathrm{x}+\mathrm{y}=3$ 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 $\text{QR}$ and $\mathrm{PQ}=\frac{2 \sqrt{2}}{3}$, the value of $9\left(x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}\right)$ is equal to _____________.

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Answer

Let $\mathrm{E}_{1}: \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1,(\mathrm{a}>\mathrm{b})$
$E_{2}: \frac{x^{2}}{c^{2}}+\frac{y^{2}}{d^{2}}=1,(c < d)$
$C: x^{2}+(y-1)^{2}=2$
Equation of tangent at $\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$
$x x_{1}+y\left(y_{1}-1\right)=\left(y_{1}+1\right)$
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, $9\left(\mathrm{x}_{1} \mathrm{y}_{1}+\mathrm{x}_{2} \mathrm{y}_{2}+\mathrm{x}_{3} \mathrm{y}_{3}\right)$
$9\left(2+\frac{5}{3} \times \frac{4}{3}+\frac{1}{3} \times \frac{8}{3}\right)$
$\Rightarrow 46$

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