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

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 60^0 with the x-axis is 32. Statement 2: The length of focal chord of a parabola y^2=4a x making an angle with the x-axis is 4acos e c^2alpha

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=4ax at a distance p from the vertex is:

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

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

A normal chord of the parabola y^2=4ax subtends a right angle at the vertex if its slope is

If a focal chord of y^2=4a x makes an angle alpha in [0,pi/4] with the positive direction of the x-axis, then find the minimum length of this focal chord.

If a focal chord of y^(2)=4ax makes an angle alphain[pi//4,pi//2] with the positive direction of the x-axis, then find the maximum length of this focal shord.

If a normal to a parabola y^2 =4ax makes an angle phi with its axis, then it will cut the curve again at an angle