Consider the ellipse

$$\frac{x^2}{4} + \frac{y^2}{3} = 1.$$

Let $$H(\alpha, 0)$$, $$0 < \alpha < 2$$, be a point. A straight line drawn through $$H$$ parallel to the $$y$$-axis crosses the ellipse and its auxiliary circle at points $$E$$ and $$F$$ respectively, in the first quadrant. The tangent 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) $$\frac{3\sqrt{3}}{2}$$

The correct option is:

The ellipse is $$\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$$ with semi-major axis $$a=2$$ and semi-minor axis $$b=\sqrt{3}$$.
Its auxiliary circle has equation $$x^{2}+y^{2}=a^{2}=4$$.

A vertical line through $$H(\alpha,0)\;(0\lt\alpha\lt2)$$ meets

• the ellipse at $$E(\alpha,y_{E})$$, where $$y_{E}=\sqrt{3\!\left(1-\frac{\alpha^{2}}{4}\right)}$$,
• the auxiliary circle at $$F(\alpha,y_{F})$$, where $$y_{F}=\sqrt{4-\alpha^{2}}$$.

The line $$OF$$ makes an angle $$\phi$$ with the positive $$x$$-axis, so

$$\tan\phi=\frac{y_{F}}{\alpha}=\frac{\sqrt{4-\alpha^{2}}}{\alpha}$$.

Squaring gives $$\alpha^{2}\tan^{2}\phi=4-\alpha^{2} \Longrightarrow \alpha^{2}(1+\tan^{2}\phi)=4 \Longrightarrow \alpha^{2}\sec^{2}\phi=4,$$ hence

$$\boxed{\alpha=2\cos\phi} \qquad(0\lt\phi\lt\frac{\pi}{2}).$$

With this, $$y_{F}=\sqrt{4-\alpha^{2}}=2\sin\phi.$$ Thus the segment $$FH$$ has length$$ FH=y_{F}=2\sin\phi.$$ Since $$H$$ and $$G$$ lie on the $$x$$-axis, $$\triangle FGH$$ is right-angled at $$H$$.

The tangent to the ellipse at $$E(\alpha,y_{E})$$ is given by the standard form $$\frac{xx_{1}}{4}+\frac{yy_{1}}{3}=1.$$ Substituting $$(x_{1},y_{1})=(\alpha,y_{E})$$ and putting $$y=0$$ (for the $$x$$-axis) yields $$\frac{x\alpha}{4}=1 \Longrightarrow x=\frac{4}{\alpha}.$$ Hence $$G\!\left(\dfrac{4}{\alpha},0\right)$$ and $$GH=\frac{4}{\alpha}-\alpha=\frac{4-\alpha^{2}}{\alpha} =\frac{4(1-\cos^{2}\phi)}{2\cos\phi} =\frac{2\sin^{2}\phi}{\cos\phi}.$$

The required area is therefore $$ \text{Area}(FGH)=\frac12\,(FH)\,(GH) =\frac12\,(2\sin\phi)\left(\frac{2\sin^{2}\phi}{\cos\phi}\right) =\boxed{\frac{2\sin^{3}\phi}{\cos\phi}}.$$

Case I: $$\phi=\frac{\pi}{4}$$
$$\sin\phi=\cos\phi=\frac1{\sqrt2},\; \text{Area}=2\left(\frac1{\sqrt2}\right)^{3}\!\!\Big/\!\left(\frac1{\sqrt2}\right)=1.$$ This matches List-II item $$Q$$. Case II: $$\phi=\frac{\pi}{3}$$
$$\sin\phi=\frac{\sqrt3}{2},\;\cos\phi=\frac12,$$ $$\text{Area}=2\left(\frac{\sqrt3}{2}\right)^{3}\!\Big/\!\left(\frac12\right) =\frac{3\sqrt3}{2},$$ which is item $$T$$. Case III: $$\phi=\frac{\pi}{6}$$
$$\sin\phi=\frac12,\;\cos\phi=\frac{\sqrt3}{2},$$ $$\text{Area}=2\left(\frac12\right)^{3}\!\Big/\!\left(\frac{\sqrt3}{2}\right) =\frac1{2\sqrt3},$$ which is item $$S$$. Case IV: $$\phi=\frac{\pi}{12}$$
$$\sin\phi=\frac{\sqrt6-\sqrt2}{4},\;\cos\phi=\frac{\sqrt6+\sqrt2}{4}.$$ Using $$\frac{2\sin^{3}\phi}{\cos\phi}$$: $$ \text{Area}=\frac{(\sqrt6-\sqrt2)^{3}}{32}\;\Big/\;\frac{\sqrt6+\sqrt2}{4} =\frac{(\sqrt6-\sqrt2)^{3}}{8(\sqrt6+\sqrt2)} =\frac{(\sqrt3-1)^{4}}{8}, $$ item $$P$$.

Combining the results:

(I) $$\to$$ (Q), (II) $$\to$$ (T), (III) $$\to$$ (S), (IV) $$\to$$ (P).

Hence the correct option is
Option C: (I) → (Q); (II) → (T); (III) → (S); (IV) → (P).