If the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ coincide with the foci of the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,then $b^2$ is equal to

Let $x^2=4ky, k>0$ be a parabola with vertex $O(0,0)$. Let $BC$ be its latus rectum. An ellipse with center $P$ on $BC$ touches the parabola at $O$,and cuts $BC$ at points $D$ and $E$ such that $BD=DE=EC$ ($B, D, E, C$ in that order). The eccentricity of the ellipse is

If a tangent to the hyperbola $x^2 - \frac{y^2}{3} = 1$ is also a tangent to the parabola $y^2 = 8x$,then the equation of such a tangent with a positive slope is:

If the common tangents to the parabola $x^2 = 4y$ and the circle $x^2 + y^2 = 4$ intersect at the point $P$,then find the square of the slope of the line.

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

Tangents are drawn from the point $P(3, 4)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$,touching the ellipse at points $A$ and $B$. The orthocenter of $\Delta PAB$ is: