Question

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

(a) 1
(b) 2
(c) 3
(d) $\frac{1}{2}$

Answer 1

(a) 1
Let (x₁, y₁) be the required point.
The slope of the given line is m.
We have
$$\begin{gathered} y^2=4x \\ \Rightarrow 2y\frac{dy}{dx}=4 \\ \Rightarrow \frac{dy}{dx}=\frac{4}{2y}=\frac{2}{y} \\ \text{Slope of the tangent =}\left(\frac{dy}{dx}\right){}_{\left(x_1,y_1\right)}\text{=}\frac{2}{y_1} \\ \text{Given:} \\ \text{Slope of the tangent =}\mathit{\text{m}} \\ \mathrm{Now}, \\ \frac{2}{y_1}\mathit{=}m\mathit{}\mathit{.}\mathit{.}\mathit{.}\left(1\right) \end{gathered}$$

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.

$$\begin{gathered} \therefore y_1=m{x}_1+1\mathrm{and}{y_1}^2=4x_1 \\ \Rightarrow x_1=\frac{y_1-1}{m}\mathrm{and}{x}_1=\frac{{y_1}^2}{4} \\ \text{So,} \\ \frac{y_1-1}{m}=\frac{{y_1}^2}{4} \\ \Rightarrow \frac{y_1-1}{\left({\displaystyle \frac{2}{y_1}}\right)}=\frac{{y_1}^2}{4}[\text{From (1)]} \\ \Rightarrow \frac{y_1\left(y_1-1\right)}{2}=\frac{{y_1}^2}{4} \\ \Rightarrow 2{y_{\mathit{1}}}^2-2y_1={y_1}^2 \\ \Rightarrow {y_1}^2-2y_1=0 \\ \Rightarrow {y_1}^2-2y_1=0 \\ \Rightarrow y_1\left(y_1-2\right)=0 \\ \Rightarrow y_1=0,2 \\ \text{So, For}{\mathit{\text{y}}}_1\text{=0,}m=\frac{2}{0}=\infty \\ \mathrm{For}{\mathit{\text{y}}}_1\text{=2,}m=\frac{2}{2}=1 \end{gathered}$$