If a tangent to the hyperbola $xy = -1$ is also a tangent to the parabola $y^2 = 8x$,then the equation of that tangent is

Let the tangent to the parabola $y^2=12x$ at the point $(3, \alpha)$ be perpendicular to the line $2x+2y=3$. Then the square of the distance of the point $(6, -4)$ from the normal to the hyperbola $\alpha^2x^2-9y^2=9\alpha^2$ at its point $(\alpha-1, \alpha+2)$ is equal to $........$.

For $0 < \theta < \pi / 2$,if the eccentricity of the hyperbola $x^2 - y^2 \operatorname{cosec}^2 \theta = 5$ is $\sqrt{7}$ times the eccentricity of the ellipse $x^2 \operatorname{cosec}^2 \theta + y^2 = 5$,then the value of $\theta$ 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:

Let $\lim_{x \to 2} \frac{(\tan(x-2))(rx^2 + (p-2)x - 2p)}{(x-2)^2} = 5$ for some $r, p \in R$. If the set of all possible values of $q$, such that the roots of the equation $rx^2 - px + q = 0$ lie in $(0, 2)$, be the interval $(\alpha, \beta]$, then $4(\alpha + \beta)$ equals :

$A$ 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: