Question

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

Options

• 1

• 2

• 3

• `1/2`

Answer 1

1

 

Let (x₁, y₁) be the required point.

The slope of the given line is m.

We have

\[y^2 = 4x\]

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

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

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

\[\text { Given }:\]

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

\[\text { Now }, \]

\[\frac{2}{y_1} = m . . . \left( 1 \right)\]

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.

\[\therefore y_1 = m x_1 + 1 \text { and } {y_1}^2 = 4 x_1 \]

\[ \Rightarrow x_1 = \frac{y_1 - 1}{m} \text { and } x_1 = \frac{{y_1}^2}{4}\]

\[So,\]

\[\frac{y_1 - 1}{m} = \frac{{y_1}^2}{4}\]

\[ \Rightarrow \frac{y_1 - 1}{\left( \frac{2}{y_1} \right)} = \frac{{y_1}^2}{4} [\text { From } (1)]\]

\[ \Rightarrow \frac{y_1 \left( y_1 - 1 \right)}{2} = \frac{{y_1}^2}{4}\]

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

\[ \Rightarrow {y_1}^2 - 2 y_1 = 0\]

\[ \Rightarrow {y_1}^2 - 2 y_1 = 0\]

\[ \Rightarrow y_1 \left( y_1 - 2 \right) = 0\]

\[ \Rightarrow y_1 = 0, 2\]

\[\text { So, For } y_1 =0,m = \frac{2}{0} = \infty \]

\[\text { For } y_1 =2,m = \frac{2}{2} = 1\]