At what angle do the curves $x^2 - y^2 = 5$ and $\frac{x^2}{18} + \frac{y^2}{8} = 1$ intersect at any common point?

Consider an ellipse $E$,a hyperbola $H$,and a parabola $P$ such that each curve has the focus at $(2, 3)$ and the corresponding directrix is $x + y - 10 = 0$. If $(\alpha, \alpha_1)$,$(\beta, \beta_1)$,and $(\gamma, \gamma_1)$ are the nearest vertices of the ellipse,hyperbola,and parabola to the given directrix respectively,then:

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

The ellipse $4x^2 + 9y^2 = 36$ and the hyperbola $4x^2 - y^2 = 4$ have the same foci and they intersect at right angles. Then,the equation of the circle passing through the points of intersection of the two conics is:

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

$A$ 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: