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 $e_1$ and $e_2$ be two distinct roots of the equation $x^2 - ax + 2 = 0$.  Let the sets  $S_1 = \{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of hyperbolas} \} = (\alpha, \beta),$ and $S_2 = \{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of an ellipse and a hyperbola, respectively} \} = (\gamma, \infty).$ Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:

The equations of the pairs of opposite sides of a parallelogram are $x^2 - 5x + 6 = 0$ and $y^2 - 6y + 5 = 0$. The equations of its diagonals are:

If two curves $x^2-4y^2=2$ and $8x^2=40-my^2$ are orthogonal to each other,then $m=$

Let the circle $C$ touch the line $x - y + 1 = 0$,have the centre on the positive $x$-axis,and cut off a chord of length $\frac{4}{\sqrt{13}}$ along the line $-3x + 2y = 1$. Let $H$ be the hyperbola $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$,whose one of the foci is the centre of $C$ and the length of the transverse axis is the diameter of $C$. Then $2\alpha^2 + 3\beta^2$ is equal to . . . . . .

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