If the foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ $(b^{2} < 16)$ and the hyperbola $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$ coincide,then the value of $b^{2}$ is

If $p$ and $q$ are the eccentricities of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate hyperbola respectively,then the area of the square (in sq. units) formed by the points of intersection of the ellipse $\frac{x^2}{p^2}+\frac{y^2}{q^2}=1$ and the pair of lines $x^2-y^2=0$ is

$A$ quadratic polynomial $y = f(x)$ with absolute term $3$ neither touches nor intersects the abscissa 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 scalar product of the vectors $\vec{OA}$ and $\vec{OB}$ is:

With one focus of the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$ as the centre,a circle is drawn which is tangent to the hyperbola with no part of the circle being outside the hyperbola. The radius of the circle is

If the normal to the rectangular hyperbola $xy = c^2$ at the point $t$ meets the curve again at $t_1$,then $t^3 t_1$ has the value equal to

An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P\left( \frac{1}{2}, 1 \right)$. Its one directrix is the common tangent nearer to the point $P$, to the circle $x^2 + y^2 = 1$ and the hyperbola $x^2 - y^2 = 1$. The equation of the ellipse in the standard form is: