Find the equation of the normal to the parabola $y^2 = 4x$ which is parallel to the line $y = 3x + 4$.

What is the condition for the points $(a, 0)$,$(at_1^2, 2at_1)$,and $(at_2^2, 2at_2)$ to be collinear?

If the equation of the directrix of a parabola is $3x + 4y + 15 = 0$ and the equation of the tangent at the vertex is $3x + 4y - 5 = 0$,then what is the length of the latus rectum?

Three normals are drawn from the point $(c, 0)$ to the curve $y^2=x$. If one of the normals is the $X$-axis,then the value of $c$ for which the other two normals are perpendicular to each other is

The point $(3,4)$ is the focus and $2x - 3y + 5 = 0$ is the directrix of a parabola. Its latus rectum is:

Let the parabola $y = x^2 + px + q$ passing through the point $(1, -1)$ be such that the distance between its vertex and the $x$-axis is minimum. Then the value of $p^2 + q^2$ is: