If a parabolic reflector is $20 \, cm$ in diameter and $5 \, cm$ deep,find the focus.

$A$ circle of radius $2$ unit passes through the vertex and the focus of the parabola $y^{2}=2x$ and touches the parabola $y=\left(x-\frac{1}{4}\right)^{2}+\alpha$,where $\alpha>0$. Then $(4\alpha-8)^{2}$ is equal to

Let the focal chord $PQ$ of the parabola $y^2=4x$ make an angle of $60^{\circ}$ with the positive $x$-axis,where $P$ lies in the first quadrant. If the circle,whose one diameter is $PS$,$S$ being the focus of the parabola,touches the $y$-axis at the point $(0, \alpha)$,then $5 \alpha^2$ is equal to :

Find the equation of the tangent to the curve $y^2 = 6x$ at the point $(2, -3)$.

The parabola with directrix $x+2y-1=0$ and focus $(1,0)$ is

The line $lx + my + n = 0$ will touch the parabola $y^2 = 4ax$,if