The centroid of the triangle formed by joining the feet of the normals drawn from any point to the parabola $y^2 = 4ax$ lies on

If the line $y = 2x + k$ is a normal to the parabola $y^2 = 4x$ at the point $(t^2, 2t)$,then:

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

Three points $O(0,0)$,$P(a, a^2)$,and $Q(-b, b^2)$ with $a > 0$ and $b > 0$ lie on the parabola $y = x^2$. Let $S_1$ be the area of the region bounded by the line $PQ$ and the parabola,and $S_2$ be the area of the triangle $OPQ$. If the minimum value of $\frac{S_1}{S_2}$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n) = 1$,then $m + n$ is equal to:

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

The point on the parabola $y^{2}=64x$ which is nearest to the line $4x+3y+35=0$ has coordinates