Let $L_{1}$ be a tangent to the parabola $y^{2}=4(x+1)$ and $L_{2}$ be a tangent to the parabola $y^{2}=8(x+2)$ such that $L_{1}$ and $L_{2}$ intersect at right angles. Then $L_{1}$ and $L_{2}$ meet on the straight line

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

The equation of the tangent to the parabola $y^2=16x$,which is perpendicular to the line $3x-4y+5=0$,is given by

The point on the parabola $2y = x^2$ which is nearest to the point $(0, 3)$ is

$A$ particle is moving in the $xy$-plane along a curve $C$ passing through the point $(3, 3)$. The tangent to the curve $C$ at the point $P$ meets the $x$-axis at $Q$. If the $y$-axis bisects the segment $PQ$,then $C$ is a parabola with

Let the locus of the mid-point of the chord through the origin $O$ of the parabola $y^{2}=4x$ be the curve $S$. Let $P$ be any point on $S$. Then the locus of the point, which internally divides $OP$ in the ratio $3:1$, is: