1. Ellipse and auxiliary circle

Given ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ so $a=2,\qquad b=\sqrt 3.$

Its auxiliary circle is $x^2+y^2=a^2=4.$

Let the vertical line through $H(\alpha,0),\qquad 0<\alpha<2$ meet:

• the ellipse at $E$ in first quadrant,
• the auxiliary circle at $F$ in first quadrant.

Thus both $E$ and $F$ have $x$-coordinate $\alpha$.

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2. Coordinates of $F$ and relation with $\phi$

Since $F$ lies on the auxiliary circle, $F=(\alpha,\sqrt{4-\alpha^2}).$

The line $OF$ makes angle $\phi$ with positive $x$-axis, so on the circle of radius $2$, $F=(2\cos\phi,2\sin\phi).$

Hence $\alpha=2\cos\phi.$

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3. Coordinates of $E$

Since $E$ lies on the ellipse and has $x=\alpha$, $\frac{\alpha^2}{4}+\frac{y_E^2}{3}=1$ so

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Using $\alpha=2\cos\phi$, $y_E=\sqrt{3(1-\cos^2\phi)}=\sqrt3\sin\phi.$

Thus $E=(2\cos\phi,\sqrt3\sin\phi).$

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4. Tangent at $E$ and point $G$

For ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1,$ tangent at $(x_1,y_1)$ is $\frac{xx_1}{4}+\frac{yy_1}{3}=1.$

At $E=(2\cos\phi,\sqrt3\sin\phi)$, $\frac{x(2\cos\phi)}{4}+\frac{y(\sqrt3\sin\phi)}{3}=1$ $\frac{x\cos\phi}{2}+\frac{y\sin\phi}{\sqrt3}=1.$

To find intersection with positive $x$-axis, put $y=0$: $\frac{x\cos\phi}{2}=1 \implies x=\frac{2}{\cos\phi}=2\sec\phi.$ So $G=(2\sec\phi,0).$

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5. Area of triangle $FGH$

Points: $F=(2\cos\phi,2\sin\phi),\quad H=(2\cos\phi,0),\quad G=(2\sec\phi,0).$

Since $GH$ lies on the $x$-axis, triangle $FGH$ is right-angled at $H$.

Its base: $GH=2\sec\phi-2\cos\phi=2(\sec\phi-\cos\phi).$

Its height: $FH=2\sin\phi.$

Therefore area $\Delta=\frac12\cdot GH\cdot FH$ $=\frac12\cdot 2(\sec\phi-\cos\phi)\cdot 2\sin\phi$ $=2\sin\phi(\sec\phi-\cos\phi).$

Now simplify: $\sec\phi-\cos\phi=\frac{1}{\cos\phi}-\cos\phi=\frac{1-\cos^2\phi}{\cos\phi}=\frac{\sin^2\phi}{\cos\phi}.$ So $\Delta=2\sin\phi\cdot \frac{\sin^2\phi}{\cos\phi}=\frac{2\sin^3\phi}{\cos\phi}.$

We now evaluate for each given $\phi$.

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6. Case (I): $\phi=\frac\pi4$

$\sin\frac\pi4=\frac{1}{\sqrt2},\qquad \cos\frac\pi4=\frac{1}{\sqrt2}.$

Hence

=2\cdot \frac{1}{2}=1.$$ So $$(\mathrm I)\to (Q).$$ --- 7. **Case (II): $\phi=\frac\pi3$** $$\sin\frac\pi3=\frac{\sqrt3}{2},\qquad \cos\frac\pi3=\frac12.$$ Thus $$\Delta=\frac{2\left(\frac{\sqrt3}{2}\right)^3}{\frac12} =4\cdot \frac{3\sqrt3}{8}=\frac{3\sqrt3}{2}.$$ So $$(\mathrm {II})\to (T).$$ --- 8. **Case (III): $\phi=\frac\pi6$** $$\sin\frac\pi6=\frac12,\qquad \cos\frac\pi6=\frac{\sqrt3}{2}.$$ Then $$\Delta=\frac{2\left(\frac12\right)^3}{\frac{\sqrt3}{2}} =\frac{1}{4}\cdot \frac{2}{\sqrt3}=\frac{1}{2\sqrt3}.$$ So $$(\mathrm{III})\to (S).$$ --- 9. **Case (IV): $\phi=\frac\pi{12}$** Use $$\tan\frac\pi{12}=2-\sqrt3.$$ Also, $$\Delta=\frac{2\sin^3\phi}{\cos\phi}=2\sin^2\phi\tan\phi.

But a cleaner form is $\Delta=2\sin\phi(\sec\phi-\cos\phi).$ Another useful simplification: $\Delta=2\sin\phi\cdot \frac{\sin^2\phi}{\cos\phi}=2\tan\phi\sin^2\phi.$

Using half-angle values is possible, but the listed option suggests expression in terms of $(\sqrt3-1)^4$. Let us compute directly.

\qquad \cos\frac\pi{12}=\frac{\sqrt6+\sqrt2}{4}.$$ So $$\sin^3\frac\pi{12}=\left(\frac{\sqrt6-\sqrt2}{4}\right)^3.$$ Then $$\Delta=\frac{2\sin^3(\pi/12)}{\cos(\pi/12)} =2\cdot \frac{(\sqrt6-\sqrt2)^3}{64}\cdot \frac{4}{\sqrt6+\sqrt2} =\frac{(\sqrt6-\sqrt2)^3}{8(\sqrt6+\sqrt2)}.$$ Since $$(\sqrt6-\sqrt2)=\sqrt2(\sqrt3-1),\qquad (\sqrt6+\sqrt2)=\sqrt2(\sqrt3+1),$$ we get $$\Delta=\frac{(\sqrt2)^3(\sqrt3-1)^3}{8\sqrt2(\sqrt3+1)} =\frac{2(\sqrt3-1)^3}{8(\sqrt3+1)} =\frac{(\sqrt3-1)^3}{4(\sqrt3+1)}.$$ Now rationalize: $$\frac{(\sqrt3-1)^3}{4(\sqrt3+1)} =\frac{(\sqrt3-1)^4}{4((\sqrt3+1)(\sqrt3-1))} =\frac{(\sqrt3-1)^4}{4(3-1)} =\frac{(\sqrt3-1)^4}{8}.$$ So $$(\mathrm{IV})\to (P).$$ --- 10. **Matching with options** We found:

(\mathrm I)\to(Q),\qquad (\mathrm {II})\to(T),\qquad (\mathrm {III})\to(S),\qquad (\mathrm {IV})\to(P).

$$This matches **Option C**. --- 11. **Comparison with stored correct answer** Stored correct answer = **C**. Our derived answer also = **C**. Hence they agree.$$