if (p1,q1) and (p2,q2) are the extremities of a focal chord of the parabola y?=4ax,then p1p2 is equal to:

  Dear student

The chord of the parabola which passes through the focus is called the focal chord.

Any chord to y² = 4ax which passes through the focus is called a focal chord of the parabola y² = 4ax. 

GIven:  y² = 4ax be the equation of a parabola and ,Let Pbe(p1,q1)=(at², 2at) and  the coordinates of the other extremity be  Q (p2,q2) of the focal chord through P are (at₁², 2at₁).

Then, PS and SQ, where S is the focus (a, 0), have the same slopes

⇒ (2at-0)/(at²- a) = (2at₁ - 0)/(at₁² - a)

⇒ tt₁² – t = t₁t² – t₁

⇒ (tt₁ + 1) (t₁ – t) = 0.

Hence t₁ = –1/t, i.e. the point Q is (a/t², –2a/t).

The extremities of a focal chord of the parabola y² = 4ax may be taken as the points t and –1/t. 
there fore P (p1,q1)=(at²,2at)
the point Q(p2,q2)= (a/t², –2a/t).
Therefore p1p2=$a{t}^2\times \frac{a}{t^2}=a^2$

Regards