Derive the equation and length of a chord of a parabola

The locus of the paraobola is y²= 4ax.

Hence, the focus of the parabola is (a,0) and directrix is x = -a.

Take the points where the focal chord intersects the parabola as (x₁,y₁) and (x₂,y₂).

Now draw perpendiculars from the points (x₁,y₁) and (x₂,y₂) on the axis of the parabola(i.e the x-axis).

Now, let distance between (x₁, y₁) and (a,0) as m and distance between (x₂,y₂) and (a,0) as n.

Let length of focal chord = c = m + n.------------(1)

Now m = distance of (x₁,y₁) from x = -a [By definition of parabola].

m = (a + a) - mcosA [Where A = theta]

or m = 2a/(1 + cosA)---------(2)

Similarly, n = distance of (x2,y2) from x = -a.

or n = (a + a) + ncosA   [Where A = theta].

or n = 2a/(1 - cosA)----------(3)

From (1), (2) and (3),

c = [2a/(1 + cosA)] + [2a/(1 - cosA)]

or c = 2a[{(1 - cosA) + (1 + cosA)}/(1 + cosA)(1 - cosA)]

or c = 2a[2/(1 - cos²A)]

or c = 4a/sin²A [As sin²A + cos²A = 1]

or c = 4acosec²A

Hence the length of the focal chord = 4acosec²(theta)