What is the equation of the normal to the curve $y^2 = 4ax$ at the point $(a, 2a)$?

The length of the chord of the parabola $y^2 = x$ which is bisected at the point $(2, 1)$ is

From a point $(d, 0)$, three normals are drawn to the parabola $y^{2} = x$. Then:

Assertion $(A)$: The curves $y^2 = 4x$ and $x^2 = -2y$ intersect at $(0,0)$ and $(2, -2)$ orthogonally.
Reason $(R)$: If the product of the slopes of the tangents drawn to two curves at their point of intersection is $-1$,then the curves are said to cut each other orthogonally. The correct option among the following is:

What is the area of the triangle inscribed in the parabola $y^2 = 4x$ whose vertices have $y$-coordinates $1, 2,$ and $4$?

For any non-zero real value of $m$,the equation of the parabola to which the line $m x-y+10+m^2=0$ is a tangent,is