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Consider the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1.$ Let $H (α, 0) 0 < α < 2$ be a point. A
straight line drawn through ‘H’ parallel to y-axis crosses the ellipse and its
auxiliary circle at E and F respectively in the first Quadrant. The tangents
to the Ellipse at E intersects positive x-axis in G. The straight line joining F
and origin makes an angle $ϕ$ with one positive x-axis, then match the
following:

Column-I Column-II
I) If $ϕ = \frac{π}{4}, a r e a o f Δ^{l e} F G H$ P) $\frac{{(\sqrt{3} - 1)}^{4}}{8}$
II) If $ϕ = \frac{π}{3}, a r e a o f Δ^{l e} F G H$ Q) 1
III) If $ϕ = \frac{π}{6}, a r e a o f Δ^{l e} F G H$ R) $\frac{3}{4}$
IV) If $ϕ = \frac{π}{12}, a r e a o f Δ^{l e} F G H$ S) $\frac{1}{2 \sqrt{3}}$
T) $\frac{3 \sqrt{3}}{2}$

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$I \to R, I I \to S, I I I \to Q, I V \to P$

$I \to R, I I \to T, I I I \to S, I V \to P$

$I \to Q, I I \to T, I I I \to S, I V \to P$

$I \to Q, I I \to S, I I I \to Q, I V \to P$

answer is C.

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Detailed Solution

Auxiliary circle is $x^{2} + y^{2} = 4$
$∴ F (2 \cos θ, 2 \sin θ), E = (2 \cos θ, \sqrt{3} \sin θ)$
Tangent at ‘E’ is $\frac{x}{2} \cos θ + \frac{y}{\sqrt{3}} \sin θ = 1$
$\begin{array}{l} ∴ G = (2 \sec θ, 0), H = (\cos θ, 0) \\ ∴ A r e a o f F G H = \frac{1}{2} \times 2 \sin θ \times (2 \sec θ - 2 \cos θ) \end{array}$

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