Sol. Let LSL ′ be the latusrectum of the parabola y 2 = 4ax. The equation of LSL ′ is x = a. Let P(at 2 , 2at) be a point on the parabola y 2 = 4ax and let PQRS be a rectangle such that P and S are on the parabola and Q and R on the latusrectum LSL ′ . Clearly, coordinates of Q are (a, 2at). From the symmetry, the coordinates of R and S are (a, –2at) and (at 2 , –2at)

respectively.

∴ PQ = a –at 2 and QR = 4at

Let A denote the area of the rectangle PQRS. Then,

A = PQ × QR

⇒ A = (a – at 2 ) 4at

⇒ A = 4a 2 (t –t 3 )

⇒ = 4a 2 (1 –3t 2 ) and = –24a 2 t

For maximum or minimum values of A, we must have

= 0

⇒ 1 –3t 2 = 0

⇒ t = ±

Clearly, < 0 for t =

Hence, area A is maximum when t = . The maximum value of area A is given by

A = 4a 2 =