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Question
Write the coordinates of the point at which the tangent to the curve y = 2x2 − x + 1 is parallel to the line y = 3x + 9 ?
Solution
Let (x1, y1) be the required point.
Slope of the given line = 3
\[\text { Since, the point lies on the curve } . \]
\[\text { Hence, } y_1 = 2 {x_1}^2 - x_1 + 1 . . . \left( 1 \right)\]
\[\text { Now, }y = 2 x^2 - x + 1 \]
\[ \therefore \frac{dy}{dx} = 4x - 1\]
\[\text { Now,} \]
\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =4 x_1 - 1\]
\[\text { Given }:\]
\[\text { Slope of the tangent = Slope of line }\]
\[ \therefore 4 x_1 - 1 = 3\]
\[ \Rightarrow x_1 = 1\]
\[\text { From (1), we get }\]
\[ y_1 = 2 - 1 + 1 = 2\]
\[ \therefore \left( x_1 , y_1 \right) = \left( 1, 2 \right)\]
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