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 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 θ with the positive x-axis.

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

Let us analyze the geometry of the given problem step-by-step.

The equation of the ellipse is given by:
$\frac{x^2}{4}+\frac{y^2}{3}=1$
Comparing this with the standard form $\frac{x^2}{a^2}$, we have:
$a^2=4\Rightarrow a=2$
$b^2=3\Rightarrow b=\sqrt{3}$

The auxiliary circle of the ellipse has the radius equal to the semi-major axis $a=2$. Its equation is:
$x^2+y^2=4$

Let $F$ be a point on the auxiliary circle in the first quadrant such that the line joining $F$ to the origin makes an angle $\theta$ with the positive x-axis. Thus, the coordinates of $F$ are:
$F(2cos\theta ,2sin\theta )$

Since the line passing through $F$ is parallel to the y-axis, the points $H$, $E$, and $F$ all share the same x-coordinate, $x=\alpha =2cos\theta$.
Since $H$ lies on the x-axis, its coordinates are:
$H(2cos\theta ,0)$

The point $E$ lies on the ellipse and has the x-coordinate $x=2cos\theta$. In the first quadrant, its y-coordinate is:
$y=bsin\theta =\sqrt{3}sin\theta$
Thus, the coordinates of $E$ are:
$E(2cos\theta ,\sqrt{3}sin\theta )$

The equation of the tangent to the ellipse at point $E(x_1,y_1)$ is:
$\frac{x{x}_1}{4}+\frac{y{y}_1}{3}=1$
Substituting $x_1=2cos\theta$ and $y_1=\sqrt{3}sin\theta$, we get:
$\frac{x(2cos\theta )}{4}+\frac{y(\sqrt{3}sin\theta )}{3}=1$
Simplifying:
$\frac{xcos\theta }{2}+\frac{ysin\theta }{\sqrt{3}}=1$

The tangent intersects the positive x-axis at point $G$. Setting $y=0$, we find:
$x=\frac{2}{cos\theta }$
So, the coordinates of $G$ are:
$G\left(\frac{2}{cos\theta },0\right)$

Now, let us calculate the area of $\Delta \mathrm{FGH}$. The vertices are:
$F(2cos\theta ,2sin\theta )$, $G\left(\frac{2}{cos\theta }\right)$, and $H(2cos\theta ,0)$.

Since the x-coordinates of $F$ and $H$ are the same, the line segment $\mathrm{FH}$ is vertical (parallel to the y-axis).
The length of the base $\mathrm{FH}$ is:
$\text{Base}=y_F-y_H=2sin\theta$

The height of the triangle is the horizontal distance from $G$ to the vertical line $x=2cos\theta$:
$\text{Height}=x_G-x_H=\frac{2}{cos\theta }-2cos\theta =2\left(\frac{1-{cos}^2\theta }{cos\theta }\right)=\frac{2{sin}^2\theta }{cos\theta }$

The area of $\Delta \mathrm{FGH}$ is:
$\text{Area}=\frac{1}{2}\times \text{Base}\times \text{Height}=\frac{1}{2}\times (2sin\theta )\times \left(\frac{2{sin}^2\theta }{cos\theta }\right)=\frac{2{sin}^3\theta }{cos\theta }=2{sin}^2\theta tan\theta$

Now, let us calculate the area for each given value of $\theta$:

Case (I): $\theta =\frac{\pi }{4}$
$\text{Area}=2{sin}^2\left(\frac{\pi }{4}\right)tan\left(\frac{\pi }{4}\right)=2\times {\left(\frac{1}{\sqrt{2}}\right)}^2\times 1=2\times \frac{1}{2}\times 1=1$
Therefore, (I) → (Q).

Case (II): $\theta =\frac{\pi }{3}$
$\text{Area}=2{sin}^2\left(\frac{\pi }{3}\right)tan\left(\frac{\pi }{3}\right)=2\times {\left(\frac{\sqrt{3}}{2}\right)}^2\times \sqrt{3}=2\times \frac{3}{4}\times \sqrt{3}=\frac{3\sqrt{3}}{2}$
Therefore, (II) → (T).

Case (III): $\theta =\frac{\pi }{6}$
$\text{Area}=2{sin}^2\left(\frac{\pi }{6}\right)tan\left(\frac{\pi }{6}\right)=2\times {\left(\frac{1}{2}\right)}^2\times \frac{1}{\sqrt{3}}=2\times \frac{1}{4}\times \frac{1}{\sqrt{3}}=\frac{1}{2\sqrt{3}}$
Therefore, (III) → (S).

Case (IV): $\theta =\frac{\pi }{12}$
We know that:
${sin}^2\left(\frac{\pi }{12}\right)=\frac{1-cos\left(\frac{\pi }{6}\right)}{2}=\frac{1-\frac{\sqrt{3}}{2}}{2}=\frac{2-\sqrt{3}}{4}$
And:
$tan\left(\frac{\pi }{12}\right)=2-\sqrt{3}$
Substituting these values into the area formula:
$\text{Area}=2\times \left(\frac{2-\sqrt{3}}{4}\right)\times (2-\sqrt{3})=\frac{{(2-\sqrt{3})}^2}{2}$
Since ${(\sqrt{3}-1)}^2=3-2\sqrt{3}+1=4-2\sqrt{3}=2(2-\sqrt{3})$, we have:
$2-\sqrt{3}=\frac{{(\sqrt{3}-1)}^2}{2}$
Therefore:
$\text{Area}=\frac{1}{2}{\left(\frac{{(\sqrt{3}-1)}^2}{2}\right)}^2=\frac{{(\sqrt{3}-1)}^4}{8}$
Therefore, (IV) → (P).