If the tangent to the curve y = x x 2 - 3 , x ∈ R , x ≠ ± 3 , at a point α , β ≠ 0 , 0 on it is parallel to the line 2 x + 6 y - 11 = 0 , then:

Step-by-step solution

The slope of a line a x + b y + c = 0 is - a b , hence the slope of the given line 2 x + 6 y - 11 = 0 is - 1 3 . Also, we know that the slope of tangent to a curve y = f x at a point x 1 ,   y 1 is d y d x x 1 ,   y 1 We have y = x x 2 - 3 Applying quotient rule for differentiation, i.e. d d x u v = v · d u d x - u · d v d x u 2 , we get d y d x = 1 · x 2 - 3 - x · 2 x x 2 - 3 2 ⇒ d y d x = x 2 - 3 - 2 x 2 x 2 - 3 2 ⇒ d y d x = - x 2 + 3 x 2 - 3 2 Also, we have ⇒ d y