The locus of the foot of the perpendicular drawn from the centre of the ellipse $x^2 + 3y^2 = 6$ to any tangent to it is:

For an ellipse with eccentricity $e = \frac{1}{2}$,the centre is at the origin. If one of its directrices is $x = 4$,then the equation of the ellipse is

Prove that the product of the lengths of the perpendiculars drawn from the points $(\sqrt{a^{2}-b^{2}}, 0)$ and $(-\sqrt{a^{2}-b^{2}}, 0)$ to the line $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$ is $b^{2}$.

If $\beta$ is one of the angles between the normals to the ellipse $x^2 + 3y^2 = 9$ at the points $(3\cos \theta, \sqrt{3} \sin \theta)$ and $(-3\sin \theta, \sqrt{3} \cos \theta)$,where $\theta \in (0, \pi/2)$,then $\frac{2 \cot \beta}{\sin 2\theta}$ is equal to

The maximum length of a chord of the ellipse $\frac{x^2}{8} + \frac{y^2}{4} = 1$,such that the eccentric angles of its extremities differ by $\frac{\pi}{2}$,is:

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 $\triangle ORT$ is $\frac{3}{2}$,then which of the following options is correct?