UPSEE Solved Model Paper 4

4x2+9y2 = 36
⇒

x2

9

−

y2

4

= 1
so, a = 3 , b = 2
In an ellipse, Focal points, F = (±√a2−b2,0) = (±√5,0)
So, F1F2 = 2√5 which is the base of the triangle PF1F2
An arbitary point P on ellipse is (a cos θ , b sin θ)
So, Height of ΔPF1F2 = b sin θ = 2 sin θ
i.e., Area =

1

2

base × height =

1

2

×2√5×2sinθ
= 2√5sinθ
= √10
⇒ sin θ =

1

√2

⇒ cos θ = ±

1

√2

And P = (a cos θ , b sin θ) = (±

3

√2

,√2)
So, option A has one of the answers.