Find the equation of the tangent to the parabola $x^2 = y$ at one of the endpoints of its latus rectum in the first quadrant.

If the tangent to the parabola $4y^2 = x$ makes an angle of $60^{\circ}$ with the $x$-axis,then find its point of contact.

Let $P$ represent the point $(3, 6)$ on the parabola $y^2 = 12x$. For the parabola $y^2 = 12x$,if $l_1$ is the length of the normal chord drawn at $P$ and $l_2$ is the length of the focal chord drawn through $P$,then $\frac{l_1}{l_2} = $

The point of intersection of the latus rectum and axis of the parabola $y^2+4x+2y-8=0$ is

$A$ parabola passing through the point $(-4, -2)$ has its vertex at the origin and the $y$-axis as its axis. The length of the latus rectum of the parabola is:

If the equation of the parabola,whose vertex is at $(5,4)$ and the directrix is $3x+y-29=0$,is $x^{2}+ay^{2}+bxy+cx+dy+k=0$,then $a+b+c+d+k$ is equal to