If 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:

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:

$A$ quadratic polynomial $y = f(x)$ with constant term $3$ neither touches nor intersects the $x$-axis and is symmetric about the line $x = 1$. The coefficient of the leading term of the polynomial is unity. $A$ point $A(x_1, y_1)$ with abscissa $x_1 = 1$ and a point $B(x_2, y_2)$ with ordinate $y_2 = 11$ are given in a Cartesian rectangular system of coordinates $OXY$ in the first quadrant on the curve $y = f(x)$,where $O$ is the origin. The vertex of the quadratic polynomial is:

An ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ and the parabola $x^2 = 4(y + b)$ are such that the two foci of the ellipse and the end points of the latus rectum of the parabola are the vertices of a square. The eccentricity of the ellipse is

The tangent at $P$ to a parabola $y^2 = 4ax$ meets the directrix at $U$ and the latus rectum at $V$. Then $\triangle SUV$ (where $S$ is the focus) :