The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide. Then,the value of $b^2$ is

If the extremities of the latus rectum having positive ordinate of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ lie on the parabola $x^2 + 2ay - 4 = 0$,then the points $(a, b)$ lie on the curve:

Find the locus of a point which moves such that its distance from the point $(0, 0)$ is twice its distance from the $y$-axis.

If two tangents drawn from a point $(\alpha, \beta)$ lying on the ellipse $25x^{2} + 4y^{2} = 1$ to the parabola $y^{2} = 4x$ are such that the slope of one tangent is four times the other,then the value of $(10\alpha + 5)^{2} + (16\beta^{2} + 50)^{2}$ equals

If the curves $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$ intersect at right angles,then $a^2 =$

Two mutually perpendicular tangents of the parabola $y^2 = 4ax$ meet the axis in $P_1$ and $P_2$. If $S$ is the focus of the parabola,then $\frac{1}{SP_1} + \frac{1}{SP_2}$ is equal to