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प्रश्न
Find the point on the curve y = 3x2 + 4 at which the tangent is perpendicular to the line whose slop is \[- \frac{1}{6}\] ?
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उत्तर
Let (x1, y1) be the required point.
Slope of the given line = \[\frac{- 1}{6}\]
∴ Slope of the line perpendicular to it = 6
\[\text { Since, the point lies on the curve } . \]
\[\text { Hence}, y_1 = 3 {x_1}^2 + 4\]
\[\text { Now,} y = 3 x^2 + 4\]
\[ \therefore \frac{dy}{dx} = 6x\]
\[\text { Now, }\]
\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =6 x_1 \]
\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \text{Slope of the given line [Given]}\]
\[ \therefore 6 x_1 = 6\]
\[ \Rightarrow x_1 = 1\]
\[\text {and }\]
\[ y_1 = 3 {x_1}^2 + 4 = 3 + 4 = 7\]
\[\text { Thus, the required point is }\left( 1, 7 \right).\]
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