If chords `lx+my=1` to the parabola `y^2= 4ax` make angle `90^@` at the origin then all chord pass through

All chords of the parabola y^(2)=4x which subtend right angle at the origin are concurrent at the point.

If a normal chord of a parabola y^(2) = 4ax subtends a right angle at the origin, then the slope of that normal chord is

A normal chord of the parabola y^2=4ax subtends a right angle at the vertex, find the slope of chord.

If two normals to a parabola y^2 = 4ax intersect at right angles then the chord joining their feet pass through a fixed point whose co-ordinates are:

If two normals to a parabola y^2 = 4ax intersect at right angles then the chord joining their feet pass through a fixed point whose co-ordinates are:

If two normals to a parabola y^2 = 4ax intersect at right angles then the chord joining their feet pass through a fixed point whose co-ordinates are:

The length of a focal chord of the parabola y^2=4ax making an angle theta with the axis of the parabola is (a> 0) is :

The focal chord of the parabola y^(2)=4ax makes an angle theta with its aixs. Show that the length of the chord will be 4acosec^(2)theta .

The normal chord of the parabola y^(2)=4ax subtends a right angle at the focus.Then the end point of the chord is

The normal chord of the parabola y^(2)=4ax subtends a right angle at the vertex.Then the length of chord is