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 $........$.

The equations of the common tangents to the two hyperbolas $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ are:

If the normal to the ellipse $\frac{x^2}{16} + \frac{y^2}{3} = 1$ at the point $\left( 2, \frac{3}{2} \right)$ is tangent to a parabola,find the equation of the parabola.

Let $e_1$ and $e_2$ be the eccentricities of the ellipse $\frac{x^2}{b^2} + \frac{y^2}{25} = 1$ and the hyperbola $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$,respectively. If $b < 5$ and $e_1 e_2 = 1$,then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:

An ellipse intersects the hyperbola $2x^2 - 2y^2 = 1$ orthogonally. The eccentricity of the ellipse is the reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes,then:
$(A)$ Equation of ellipse is $x^2 + 2y^2 = 2$
$(B)$ The foci of ellipse are $(\pm 1, 0)$
$(C)$ Equation of ellipse is $x^2 + 2y^2 = 4$
$(D)$ The foci of ellipse are $(\pm \sqrt{2}, 0)$

$A$ 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: