#### Question

Find the coordinates of the point on the curve y^{2} = 3 − 4x where tangent is parallel to the line 2x + y− 2 = 0 ?

#### Solution

Let (*x*_{1}*, y*_{1}) be the required point.

Slope of the given line = \[-\] 2

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

\[\text { Hence,} {y_1}^2 = 3 - 4 x_1 . . . \left( 1 \right)\]

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

\[ \Rightarrow 2y\frac{dy}{dx} = - 4\]

\[ \therefore \frac{dy}{dx} = \frac{- 4}{2y} = \frac{- 2}{y}\]

\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{- 2}{y_1}\]

\[\text { Given }:\]

\[\text { Slope of the tangent = Slope of the line }\]

\[ \Rightarrow \frac{- 2}{y_1} = - 2\]

\[ \Rightarrow y_1 = 1\]

\[\text { From (1), we get }\]

\[1 = 3 - 4 x_1 \]

\[ \Rightarrow - 2 = - 4 x_1 \]

\[ \Rightarrow x_1 = \frac{1}{2}\]

\[ \therefore \left( x_1 , y_1 \right) = \left( \frac{1}{2}, 1 \right)\]