Show that the focal chord, of parabola `y^2 = 4ax`, that makes an angle `alpha` with the x-axis is of length `4a cosec^2 alpha`.

Statement 1: The length of focal chord of a parabola y^(2)=8x making on an angle of 60^(@) with the x-axis is 32. Statement 2: The length of focal chord of a parabola y^(2)=4ax making an angle with the x -axis is 4a cos ec^(2)alpha

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:

length of the focal chord of the parabola y^(2)=8x making an angle 60^(@) with the x- axis is l, then (3l)/(16)=

If the normal chord of the parabola y^(2)=4 x makes an angle 45^(@) with the axis of the parabola, then its length, is

Length of the normal chord of the parabola y^(2)=4x which makes an angle of (pi)/(4) ,with the x -axis is k sqrt(2) ,then k=

Length of the focal chord of the parabola y^(2)=4ax at a distance p from the vertex is:

The length of a focin chord of the parabola y2=4ax making an angle with the axis of the parabola is (A)4a cos ec^(2)theta(C)a cos ec2(B)4a sec2 theta(D) none of these

The length of a focal chord of the parabola y^(2) = 4ax at a distance b from the vertex is c. Then

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

Length of the shortest normal chord of the parabola y^(2)=4ax is