QJEEAdv16Ma9caa: JEE Advanced 2016 Mathematics - Class 11 JEEAdv16Ma9caa Solution: The circle with centre at , intersects the parabola at the…

In case of option (A)

x² + y² = 3 and x² = 2y

Solving we get $P(\sqrt 2 ,1)$

Equation of tangent at P on the circle

$x\sqrt 2 + y = 3$ $\therefore$ $\tan \alpha = - \sqrt 2$

Again, ${{{Q_2}{R_2}} \over {{Q_2}M}} = \sin \left( {\alpha - {\pi \over 2}} \right)$

or, ${{2\sqrt 3 } \over {{Q_2}M}} = - \cos \alpha = - {1 \over {\sqrt 3 }}$ [ $\because$ $\tan \alpha = - \sqrt 2$ ]

or, ${Q_2}M = 6$

Similarly, ${Q_3}M = 6$

$\therefore$ ${Q_2}{Q_3} = 12$

In case of option (B)

${{M{R_2}} \over {{Q_2}{R_2}}} = \tan \left( {\alpha - {\pi \over 2}} \right)$

or, ${{M{R_2}} \over {2\sqrt 3 }} = - \tan \alpha = \sqrt 2$

or, $M{R_2} = 2\sqrt 6$

Similarly, $M{R_3} = 2\sqrt 6$ $\therefore$ ${R_2}{R_3} = 4\sqrt 6$

In case of option (C)

OP is the perpendicular drawn from O to R₂R₃

$\therefore$ area of $\Delta O{R_2}{R_3} = {1 \over 2} \times OP \times {R_2}{R_3} = {1 \over 2} \times \sqrt 3 \times 4\sqrt 6 =6\sqrt 2$

In case of option (D)

PT is perpendicular drawn from P to Q₂Q₃

$\therefore$ area of $\Delta P{Q_2}{Q_3} = {1 \over 2} \times PT \times {Q_2}{Q_3} = {1 \over 2} \times \sqrt 2 \times 12 = 6\sqrt 2$

Therefore, (A), (B), (C) are correct options.