Question

Find the point on the curve y = 3x² + 4 at which the tangent is perpendicular to the line whose slop is \[- \frac{1}{6}\]  ?

Answer 1

Let (x₁, y₁) be the required point.
Slope of the given line = \[\frac{- 1}{6}\]

∴ Slope of the line perpendicular to it = 6

\[\text { Since, the point lies on the curve } . \]

\[\text { Hence}, y_1 = 3 {x_1}^2 + 4\]

\[\text { Now,} y = 3 x^2 + 4\]

\[ \therefore \frac{dy}{dx} = 6x\]

\[\text { Now, }\]

\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =6 x_1 \]

\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \text{Slope of the given line [Given]}\]

\[ \therefore 6 x_1 = 6\]

\[ \Rightarrow x_1 = 1\]

\[\text {and }\]

\[ y_1 = 3 {x_1}^2 + 4 = 3 + 4 = 7\]

\[\text { Thus, the required point is }\left( 1, 7 \right).\]