Consider the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. Let $S(p, q)$ be a point in the tirst quadrant such that $\frac{p^2}{9}+\frac{q^2}{4}>1$. I wo 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?

The centre of the ellipse $4{x^2} + 9{y^2} - 16x - 54y + 61 = 0$ is

Let $E$ be the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$. For any three distinct points $P, Q$ and $Q^{\prime}$ on $E$, let $M(P, Q)$ be the mid-point of the line segment joining $P$ and $Q$, and $M \left( P , Q ^{\prime}\right)$ be the mid-point of the line segment joining $P$ and $Q ^{\prime}$. Then the maximum possible value of the distance between $M ( P , Q )$ and $M \left( P , Q ^{\prime}\right)$, as $P, Q$ and $Q^{\prime}$ vary on $E$, is. . . . .

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

A wall is inclined to the floor at an angle of $135^{\circ}$. A ladder of length $l$ is resting on the wall. As the ladder slides down, its mid-point traces an arc of an ellipse. Then, the area of the ellipse is

Let $x^2=4 k y, k>0$ be a parabola with vertex $A$. Let $B C$ be its latusrectum. An ellipse with centre on $B C$ touches the parabola at $A$, and cuts $B C$ at points $D$ and $E$ such that $B D=D E=E C(B, D, E, C$ in that order). The eccentricity of the ellipse is