$ y+9=m(x-4) $
$ \therefore \quad A=\left(\frac{9+4 m}{m}, 0\right) $
$ \quad B=(0,-9-4 m) $
$ \therefore \quad O A+O B=\frac{9+4 m}{m}+9+4 m$
$ \because \mathrm{m}>0 $
$ =13+\frac{9}{\mathrm{~m}}+4 \mathrm{~m} $
$ \because \frac{4 \mathrm{~m}+\frac{9}{\mathrm{~m}}}{2} \geq \sqrt{36} \Rightarrow 4 \mathrm{~m}+\frac{9}{\mathrm{~m}} \geq 12 $
$ \therefore \mathrm{OA}+\mathrm{OB} \geq 25$
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$\frac{x^2}{4}+\frac{y^2}{3}=1$
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.
| $List-I$ | $List-II$ |
| If $\phi=\frac{\pi}{4}$, then the area of the triangle $F G H$ is | ($P$) $\frac{(\sqrt{3}-1)^4}{8}$ |
| If $\phi=\frac{\pi}{3}$, then the area of the triangle $F G H$ is | ($Q$) $1$ |
| If $\phi=\frac{\pi}{6}$, then the area of the triangle $F G H$ is | ($R$) $\frac{3}{4}$ |
| If $\phi=\frac{\pi}{12}$, then the area of the triangle $F G H$ is | ($S$) $\frac{1}{2 \sqrt{3}}$ |
| ($T$) $\frac{3 \sqrt{3}}{2}$ |
The correct option is: