Question

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 points E and F respectively, in the first quadrant. The tangents to the ellipse at the point E intersects the positive x-axis at a point G. Suppose the straight line joining F and the origin makes an angle \(\phi\) with the positive x-axis.
 

List-I		List-II
(I)	If \(\Phi = \frac{\pi}{4}\), then the area of the triangle FGH is	P	\(\frac{(\sqrt{3}-1)^4}{8}\)
(II)	If \(\Phi = \frac{\pi}{3}\), then the area of the triangle FGH is	Q	1
(III)	If \(\Phi = \frac{\pi}{6}\), then the area of the triangle FGH is	R	\(\frac{3}{4}\)
(IV)	If \(\Phi = \frac{\pi}{12}\), then the area of the triangle FGH is	S	\(\frac{1}{2\sqrt{3}}\)
		T	\(\frac{3\sqrt{3}}{2}\)

• A. (I) → (R); (II) → (S); (III) → (Q); (IV) → (P)
• B. (I) → (R); (II) → (T); (III) → (S); (IV) → (P)
• C. (I) → (Q); (II) → (T); (III) → (S); (IV) → (P)
• D. (I) → (Q); (II) → (S); (III) → (Q); (IV) → (P)

Answer 1

The Correct Option is C

Equation of auxiliary circle \(x^2 + y^2 = 4\)

Let F be \((2 cos \theta, 2 sin \theta)\)

 E is \(( 2cos\theta , \sqrt{3} sin \theta)\)

Equation of tangent at E \(x \cos{\frac{\theta}{2}}+ y \sin{\frac{\theta}{\sqrt{3}}}=1\)

It cuts x-axis at \((2 sec \theta, 0)\)

Therefore, G is \((2 sec \theta, 0)\)

H is \((2 cos \theta, 0)\) and F\((2 cos \theta, 2 sin \theta)\)

Area of \(\triangle FGH\) is \(\frac{1}{2}\times2sin\theta(2 sec\theta-2cos\theta)\)\(=2 sin\theta(sec\theta-cos\theta)\)

\((I) \quad f\left(\frac{\pi}{4}\right)=1\)
\((II) \quad f\left(\frac{\pi}{3}\right)=\frac{3\sqrt{3}}{2}\)

\((III) \quad f\left(\frac{\pi}{6}\right)=\frac{1}{2\sqrt{3}}\)

\((IV) \quad f\left(\frac{\pi}{12}\right)=2(2-\sqrt{3})\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^2=\frac{(4-2\sqrt{3})(\sqrt{3}-1)^2}{8}=\frac{(\sqrt{3}-1)^4}{8}\)

So, the correct option is (C): (I) → (Q); (II) → (T); (III) → (S); (IV) → (P)