Find the point on the parabola `y^2= 4ax (a > 0)` which forms a triangle of area `3a^2` with the vertex and focus of the parabola.

The focus of the parabola y^(2) = - 4ax is :

Find the locus of the middle points of the chords of the parabola y^(2)=4ax which subtend a right angle at the vertex of the parabola.

Find the vertex and focus of the given parabola x^(2)+4y=0

An equilateral triangle is inscribed in the parabola y^(2)=4ax, such that one vertex of this triangle coincides with the vertex of the parabola.Then find the side length of this triangle.

Find the locus of the mid-points of the chords of the parabola y^(2)=4ax which subtend a right angle at vertex of the parabola.

An isosceles triangle is inscribed in the parabola y^2 = 4ax with its base as the line joining the vertex and positive end of the latus rectum of the parabola. If (at^2, 2at) is the vertex of the triangle then

If the parabola y^(2) = 4ax passes through the point (4, 1), then the distance of its focus the vertex of the parabola is

Statement 1: Normal chord drawn at the point (8,8) of the parabola y^(2)=8x subtends a right angle at the vertex of the parabola.Statement 2: Every chord of the parabola y^(2)=4ax passing through the point (4a,0) subtends a right angle at the vertex of the parabola.