The angle of intersection between the curves $x^2 = 4(y + 1)$ and $x^2 = -4(y + 1)$ is

The length of the latus rectum of the parabola $(x-2)^2+(y-3)^2=\frac{1}{25}(3x-4y+7)^2$ is

Consider the parabola $y^2=8x$. Let $\Delta_1$ be the area of the triangle formed by the endpoints of its latus rectum and the point $P\left(\frac{1}{2}, 2\right)$ on the parabola,and $\Delta_2$ be the area of the triangle formed by the intersection points of the tangents drawn at $P$ and at the endpoints of the latus rectum. Then $\frac{\Delta_1}{\Delta_2}$ is

If $P(h, k)$ is a point on the parabola $x = 4y^2$ which is nearest to the point $Q(0, 33)$,then the distance of $P$ from the directrix of the parabola $y^2 = 4(x + y)$ is equal to:

What is the focus of the parabola $4y^2 + 12x - 20y + 67 = 0$?

$P$ and $Q$ are two distinct points on the parabola,$y^2 = 4x$,with parameters $t$ and $t_1$ respectively. If the normal at $P$ passes through $Q$,then the minimum value of $t_1^2$ is