The line y = mx + 1 is a tangent to the curve y2 = 4x, if the value of m is ________________ .

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1 Let (x1, y1) be the required point. The slope of the given line is m. We have y^2 = 4x \Rightarrow 2y dy/dx = 4 \Rightarrow dy/dx = $\frac{4}{2}$y = 2/y \text { Slope of the tangent } =\left( dy/dx \right) {}_\left( x_1 , y_1 \right) =2/y_1 \text { Given }: \text { Slope of the tangent }=m \text { Now }, 2/y_1 = m . . . \left( 1 \right) Because the given line is a tangent to the given curve at point (x1, y1), this point lies on both the line and the curve. \therefore y_1 = m x_1 + 1 \text { and } {y_1}^2 = 4 x_1 \Rightarrow x_1 = y_1 - 1/m \text { and } x_1 = \frac{{y_1}^2}{4} So, y_1 - 1/m = \frac{{y_1}^2}{4} \Rightarrow \frac{y_1 - 1}{\left( 2/y_1 \right)} = \frac{{y_1}^2}{4} [\text { From } (1)] \Rightarrow \frac{y_1 \left( y_1 - 1 \right)}{2} = \frac{{y_1}^2}{4} \Rightarrow 2 {y_1}^2 - 2 y_1 = {y_1}^2 \Rightarrow {y_1}^2 - 2 y_1 = 0 \Rightarrow {y_1}^2 - 2 y_1 = 0 \Rightarrow y_1 \left( y_1 - 2 \right) = 0 \Rightarrow y_1 = 0, 2 \text { So, For } y_1 =0,m = $\frac{2}{0}$ = \infty \text { For } y_1 =2,m = $\frac{2}{2}$ = 1

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