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

The points of intersection of the curves whose parametric equations are $x = t^2 + 1, y = 2t$ and $x = 2s, y = \frac{2}{s}$ is given by

The triangle $PQR$ of area $A$ is inscribed in the parabola $y^2 = 4ax$ such that the vertex $P$ lies at the vertex of the parabola and the base $QR$ is a focal chord. The modulus of the difference of the ordinates of the points $Q$ and $R$ is:

$A$ line perpendicular to the $X$-axis cuts the circle $x^2+y^2=9$ at $A$ and the ellipse $4x^2+9y^2=36$ at $B$ such that $A$ and $B$ lie in the same quadrant. If $\theta$ is the greatest acute angle between the tangents drawn to the curves at $A$ and $B$,then $\tan \theta=$

The values of $\lambda$,for which the point $(\lambda, \lambda-2)$ lies inside the ellipse $4x^2+9y^2=36$ and outside the parabola $y^2=x$,satisfy:

Let $\alpha, \beta$ be the roots of the quadratic equation $x^2 + px + p^3 = 0$ $(p \neq 0)$. If $(\alpha, \beta)$ is a point on the parabola $y^2 = x$,then the roots of the quadratic equation are: