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
(II)	If $\Phi = \frac{\pi}{3}$, then the area of the triangle FGH is
(III)	If $\Phi = \frac{\pi}{6}$, then the area of the triangle FGH is
(IV)	If $\Phi = \frac{\pi}{12}$, then the area of the triangle FGH is
	T

• 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

Answer: (C)

(I) → (Q); (II) → (T); (III) → (S); (IV) → (P)

Answer 2

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 $\frac{1}{2}\times2sin\theta(2 sec\theta-2cos\theta)$
$=2 sin\theta(sec\theta-cos\theta)$

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

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

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

$(I) → (Q); (II) → (T); (III) → (S); (IV) → (P)$