The radius of the circle passing through the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ and having its center at $(0, 3)$ is:

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

If $P$ is any point on the ellipse $16x^2 + 25y^2 = 400$ with foci $S$ and $S'$, and the area of $\Delta PSS'$ is $9 \text{ square units}$, then the abscissa of point $P$ is:

If the normal at one end of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ passes through one end of the major axis,then:

Let for two distinct values of $p$ the lines $y=x+p$ touch the ellipse $E: \frac{x^2}{16} + \frac{y^2}{9} = 1$ at the points $A$ and $B$. Let the line $y = x$ intersect $E$ at the points $C$ and $D$. Then the area of the quadrilateral $ABCD$ is equal to

The line $x=m^2$ meets the ellipse $9x^2+y^2=9$ at real and distinct points if and only if