Find a point on the curve y = 3x^2 + 4 at which the tangent is perpendicular to the line whose slope is -1/6

Question

Find a point on the curve y = 3x² + 4 at which the tangent is perpendicular to the line whose slope is \(-\frac{1}{6}\).

Answer 1

Given:

The curve y = 3x² + 4 and the Slope of the tangent is \(\frac{-1}{6}\)

y = 3x² + 4

Differentiating the above w.r.t x

⇒ \(\frac{dy}{dx}\) = 2 x 3x^{2 – 1} + 0

⇒ \(\frac{dy}{dx}\) = 6x ...(1)

Since, tangent is perpendicular to the line,

\(\therefore\) The Slope of the normal = \(\frac{-1}{\text{The Slope of the tangent}}\)

i.e, \(\frac{-1}{6}\) = \(\frac{-1}{6x}\)

⇒   \(\frac{1}{6}\) = \(\frac{1}{6x}\)

⇒ x = 1

Substituting x = 1 in y = 3x² + 4,

⇒ y = 3(1)² + 4

⇒ y = 3 + 4

⇒ y = 7

Thus, the required point is (1,7).