Parabola

On solving x2+y2=3 and x^2=2y we get point (√2,1)
Equation of tangent at P
√2.x+y=3
Let Q2 be (0,k) and radius is 2√3
∴|

√2(0)+k−3

√2+1

|=2√3
∴k=9,−3
Q2(0,9) and Q3(0,−3)
hence Q2,Q3=12
R2,R3 is internal common tangent of circle C2 and C3
∴R2,R3=√(Q2,Q3)2−(2√3+2√3)2
=√122−48=√96=4√6
Perpendicular distance of origin O from R2,R3 is equal to radius of circle C1=√3

Perpendicular Distance of P from Q2Q3=√2
∴Area of ΔPQ2Q3=

1

2

×12×√2=6√2