Let a tangent to the curve $y^2 = 24x$ meet the curve $xy = 2$ at the points $A$ and $B$. Then the midpoints of such line segments $AB$ lie on a parabola with the

The normal at $\left( 2, \frac{3}{2} \right)$ to the ellipse $\frac{x^2}{16} + \frac{y^2}{3} = 1$ touches a parabola,whose equation is

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$,with $y_1 < 0$ and $y_2 < 0$,be the endpoints of the latus rectum of the ellipse $x^2 + 4y^2 = 4$. The equations of the parabolas with latus rectum $PQ$ are:
$(A) x^2 + 2\sqrt{3}y = 3 + \sqrt{3}$
$(B) x^2 - 2\sqrt{3}y = 3 + \sqrt{3}$
$(C) x^2 + 2\sqrt{3}y = 3 - \sqrt{3}$
$(D) x^2 - 2\sqrt{3}y = 3 - \sqrt{3}$

If the curves $2x^2 + ky^2 = 30$ and $3y^2 = 28x$ cut each other orthogonally,then $k=$

In the figure, Statement $I$: when $\alpha > \beta \ge 0$, the section is a hyperbola. Statement $II$: when $\beta > 90^\circ$, the section is an ellipse.

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 $........$.