Find the equation of the normal to the parabola $y^2 = 16x$ having a slope of $-1/4$.

If two tangents to the parabola $y^2=8x$ meet the tangent at its vertex in $M$ and $N$ such that $MN=4$,then the locus of the point of intersection of those two tangents is

The ordinates of the points $P$ and $Q$ on the parabola with focus $(3,0)$ and directrix $x = -3$ are in the ratio $3:1$. If $R(\alpha, \beta)$ is the point of intersection of the tangents to the parabola at $P$ and $Q$,then $\frac{\beta^2}{\alpha}$ is equal to $.............$.

Normals are drawn from the point $P(8,0)$ to the parabola $y^2=12x$. If $\theta$ is the acute angle between two non-horizontal normals among them,then $\tan \theta=$

Let $PQ$ and $RT$ be two focal chords of the parabola $y^2=16x$. If $P=(4,8)$ and $R=(16,16)$,then the length of $QT$ is:

From the point $(-1, 2)$, tangents are drawn to the parabola $y^2 = 4x$. The area of the triangle formed by the chord of contact and the tangents is: (in $\sqrt{2}$)