Consider the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. Let $S(p, q)$ be a point in the first quadrant such that $\frac{p^2}{9}+\frac{q^2}{4}>1$. Two tangents are drawn from $S$ to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point $T$ in the fourth quadrant. Let $R$ be the vertex of the ellipse with positive $x$-coordinate and $O$ be the center of the ellipse. If the area of the triangle $\triangle O R T$ is $\frac{3}{2}$, then which of the following options is correct?
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 $F G H$ is | (P) $\frac{(\sqrt{3}-1)^{4}}{8}$ |
| (II) If $\phi=\frac{\pi}{3}$, then the area of the triangle $F G H$ is | (Q) 1 |
| (III) If $\phi=\frac{\pi}{6}$, then the area of the triangle $F G H$ is | (R) $\frac{3}{4}$ |
| (IV) 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:
Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve
The orthocentre of the triangle $${F_1}MN$$ is