JEE Advanced 2023 — Paper 1 — Question 2

$y=m x \pm \sqrt{6 m^{2}+3} \quad$ (eq. of tangent for ellipse) $y=m x+\frac{3}{m} \quad$ (eq. of tangent for parabola)

$\Rightarrow \frac{3}{\mathrm{~m}}=\sqrt{6 \mathrm{~m}^{2}+3}$

$\Rightarrow \frac{9}{\mathrm{~m}^{2}}=6 \mathrm{~m}^{2}+3$

$\Rightarrow 3=2 \mathrm{~m}^{4}+\mathrm{m}^{2}$

$\Rightarrow 2 \mathrm{~m}^{4}+\mathrm{m}^{2}-3=0$

$\Rightarrow 2 \mathrm{~m}^{4}+3 \mathrm{~m}^{2}-2 \mathrm{~m}^{2}-3=0$

$\Rightarrow m^{2}\left(2 m^{2}+3\right)-1\left(2 m^{2}+3\right)=0$ $\Rightarrow \mathrm{m}=1,-1$

$\Rightarrow$ Equations of tangents are $\mathrm{y}=\mathrm{x}+3$ and $y=-x-3$

$\Rightarrow$ Point of intersection $=(-3,0)$

Eq. of l_{1} \rightarrow $$\mathrm{T}=0 (chord of contact for ellipse) $\frac{-3 x}{6}=1 \Rightarrow x=-2$

Eq. of l_{2} \rightarrow $$\mathrm{T}=0 (chord of contact for parabola) $12\left(\frac{\mathrm{x}-3}{2}\right) =0 \Rightarrow \mathrm{x}=3$

$\Rightarrow$ area of quadrilateral $\mathrm{A}_{1} \mathrm{~A}_{2} \mathrm{~A}_{3} \mathrm{~A}_{4}=\frac{1}{2}(2+12) \times 5=35$ sq. units