If ∀h∈R−{0}, two distinct tangents can be drawn from the points (2+h,3h−1) to the curve y=x3−6x2−a+bx, then ab is
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Solution
One of the tangents must pass through the the point of inflection of the curve. y′′=0⇒x=2 P(2,2b−a−16) Equation of tangent at P is y=(b−12)x−a+8 ...(1) Let Q = (2 + h, 3h - 1) Locus of Q →3x−y=7 ...(2) (1) and (2) must be identical. a = 15, b = 15