Show that the normals at the points `(4a, 4a)` & at the upper end of the latus rectum of the parabola `y^2 = 4ax` intersect on the same parabola.

The end points of latus rectum of the parabola x ^(2) =4ay are

The end points of the latus rectum of the parabola x ^(2) + 5y =0 is

The ends of the latus rectum of the parabola (x-2)^(2)=-6(y+1) are

The normal to y^(2)=4a(x-a) at the upper end of the latus rectum is

The one end of the latus rectum of the parabola y^(2) - 4x - 2y – 3 = 0 is at

One end of latus rectum of the parabola (x+2)^(2)=-4(y+3) is

The point of intersection of the tangents at the ends of the latus rectum of the parabola y^(2)=4x is

Find the measure of the angle subtended by the latus rectum of the parabola y^(2) = 4ax at the vertex of the parabola .

Statement-1: Point of intersection of the tangents drawn to the parabola x^(2)=4y at (4,4) and (-4,4) lies on the y-axis. Statement-2: Tangents drawn at the extremities of the latus rectum of the parabola x^(2)=4y intersect on the axis of the parabola.

Prove that the normals at the points (1,2) and (4,4) of the parabola y^(2)=4x intersect on the parabola