Let T _ 1 and T _ 2 be two distinct common tangents to the ellipse...

Given,

Equation of ellipse

$E: \dfrac{{x}^{2}}{6} + \dfrac{{y}^{2}}{3} = 1$,

Now equation of tangent with slope ${m}_{1}$ will be :

${T}_{1}: y = {m}_{1}x \pm \sqrt{6{m}_{1}^{2} + 3}$

And equation of parabola,

$P: {y}^{2} = 12x$,

So, equation of tangent with slope ${m}_{2}$ will be:

$y = {m}_{2}x + \dfrac{3}{{m}_{2}}$

Now for common tangent

$m = {m}_{1} = {m}_{2}, \pm \sqrt{6{m}_{1}^{2} + 3} = \dfrac{3}{{m}_{2}}$

$⇒ m = \pm 1$

Hence, equation of common tangents will be,

$y = x + 3$ and $y = – x - 3$

Now we know that,

Point of contact for parabola is \left(\dfrac{a}{{m}^{2}}, \dfrac{2a}{m}\right)$</p><p>$⇒ {A}_{1} ≡ \left(3, 6\right), {A}_{4}\left(3 - 6\right)$</p><p>Now let${A}_{2}\left({x}_{1}, {y}_{1}\right)$</p><p>So, equation of tangent to ellipse$E$at point${A}_{2}\left({x}_{1}, {y}_{1}\right)$is given by,</p><p>$\dfrac{x{x}_{1}}{6} + \dfrac{y{y}_{1}}{3} = 1$</p><p>Now comparing above equation with$y = x + 3$,</p><p>We get,$\dfrac{- {x}_{1}}{6} = \dfrac{{y}_{1}}{3} = \dfrac{1}{3}$</p><p>$⇒ {A}_{2} ≡ \left(\dfrac{- 6}{3}, \dfrac{3}{3}\right) ≡ \left(- 2, 1\right)$</p><p>Now${A}_{3}$is mirror image of${A}_{2}$in$x$-axis </p><p>$⇒ {A}_{3}\left(- 2, - 1\right)$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/1e9359d0-02bf-4a5f-b6b6-38c4c60027f2.png" style="width: 300px; height: 180px;" /></p><p>Now finding the intersection point of${T}_{1} = 0$and${T}_{2} = 0$, we get \left(- 3, 0\right)$

So, Area of quadrilateral ${A}_{1}{A}_{2}{A}_{3}{A}_{4} = \dfrac{1}{2}\left(12 + 2\right) \times 5 = 35$ square units