The circle C 1 : x 2 + y 2 = 3 , with centre at O, intersects the parabola x 2 = 2 y at the point P in the…

1. $Q_2{Q}_3=12$
2. $R_2{R}_3=4\sqrt{6}$
3. area of the triangle $O{R}_2 R_3$ is $6\sqrt{2}$
4. area of the triangle $P{Q}_2{Q}_3$ is $4\sqrt{2}$

Solution

On solving $x^2+y^2=3$ and $x^2=2y$ we get point $P\left(\sqrt{2}, 1\right)$
Equation of tangent at P
$\sqrt{2} x+y=3$
Let $Q_2$ be (0, k) and radius is $2\sqrt{3}$
$\therefore$ $\left|\frac{\sqrt{2} \left(0\right)+k-3}{\sqrt{2+1}}\right|=2\sqrt{3}$
$\therefore$ $k=9, -3$ $Q_3$
$Q_2 \left(0, 9\right)$ and $Q_3$ $(0,-3)$ $\{\begin{array}{c}{C}_2:{\left(x-0\right)}^2+{\left(y-9\right)}^2=12\\ C_3:{\left(x-0\right)}^2+{\left(y+3\right)}^2=12\end{array}$
Hence $Q_2 Q_3=12$

${\Rightarrow a}^2+12=36$
$\Rightarrow a=2\sqrt{6}$
${\mathrm{R}}_2{\mathrm{R}}_3=2\mathrm{a}=4\sqrt{6}$
Perpendicular distance of origin O from $R_2{R}_3$ is equal to distance of O from tangent $\sqrt{2}x+y=3$ which is same as radius of circle $C_1= \sqrt{3}$
Hence area of $\mathrm{\Delta }O{R}_2{R}_3=\frac{1}{2}\times \left(R_2{R}_3\right) \sqrt{3}=\frac{1}{2}. 4\sqrt{6} . \sqrt{3}=6\sqrt{2}$
Perpendicular Distance of P from $Q_2{Q}_3= \sqrt{2}$
$\therefore$ Area of $\mathrm{\Delta }P{Q}_2{Q}_3=\frac{1}{2}\times 12\times \sqrt{2}=6\sqrt{2}$

Asked in: JEE Advanced 2016 (Paper 1)

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