(a) Line y = 4x - 5 → slope of line m = 4 ...(i) curve y2=ax3 + b ∴ differentiating w.r.t. ‘x’ 2y
dy
dx
=3ax2
dy
dx
=
3ax2
2y
= slope of tangent ∴
dy
dx
|(2,3)=
3a×4
2×3
= 2a from(i) and (ii), we get 4 = 2a ⇒ a = 2 Since, (2,3) is a point on the curve: y2=ax3 + b. ∴ (3)2=2(2)3 + b ⇒ b = - 7 ∴ 7a + 2b = 7 x 2 + 2 (-7) = 0