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
Hence area of ΔOR2R3=
1
2
×(R2,R3)√3=
1
2.4
√6.√3=6√2
Perpendicular Distance of P from Q2Q3=√2 ∴Area of ΔPQ2Q3=