The equation of the ellipse whose one vertex is $(0, 7)$ and the directrix is $y = 12$ is:

The locus of the point of intersection of two perpendicular tangents to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is:

If the straight line $8x + 3\sqrt{2}y = 36$ touches the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 2$ at $(a, b)$,then $a + \sqrt{2}b =$

$A$ 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 $S(1,0)$ and $S^{\prime}(0,1)$ be the foci of an ellipse such that $SP+S^{\prime} P=2$ for any point $P$ on the ellipse. If $A(x_1, y_1)$ and $A^{\prime}(x_2, y_2)$ are the end points of the major axis of this ellipse,then $x_1+x_2=$

The normal at a variable point $P$ on an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ of eccentricity $e$ meets the axes of the ellipse in $Q$ and $R$. Then the locus of the mid-point of $QR$ is a conic with an eccentricity $e'$ such that: