Sol. Let the lines y = β and x = α meet y = x at Q and R respectively. The coordinates of Q and R are ( β, β ) and ( α, α ) respectively.

Let A denote the area of trapezium PQOS. Then,

A = (PQ + OS) × PS

⇒ A = ( α – β + α ) × β

⇒ A = (2 α – β ) β

Since ( α, β ) lies on 4x 2 + 3y 2 = 12 or,

+ = 1. So, let α = cos θ & β = 2 sin θ .

∴ A = (2 cos θ –2 sin θ ). 2 sin θ

⇒ A = 2 ( sin θ cos θ – sin 2 θ)

⇒ = 2 ( cos 2 θ – sin 2 θ – 2 sin θ cos θ )

⇒ = 2( cos 2 θ – sin 2 θ )

For maximum or minimum values of A, we must have = 0

⇒ cos 2 θ – sin 2 θ = 0

⇒ tan 2 θ =

⇒ 2 θ =

⇒ θ =

Now,

= 2 (–2 sin 2 θ –2 cos 2 θ )

Clearly, < 0 for θ =

Hence, A is maximum when θ = . For this value of θ , we have

α = cos = and β = 2 sin = 1

Hence, the required point is (3/2, 1)