# Choose the correct answer in the line $$y = mx + 1$$ is a tangent to the curve $$y^2 = 4x$$ if the value of $$m$$ is

$(A)\; 1 \quad (B)\; 2 \quad (C)\; 3 \quad (D)\;\frac{1}{2}$

Toolbox:
• $\large\frac{d}{dx}$$(x^n)=nx^{n-1} • Equation of tangent y=mx+1 Step 1: The equation of the curve is y^2=4x Differentiating with respect to x 2y\large\frac{dy}{dx}$$=4$
$\large\frac{dy}{dx}=\frac{4}{2y}$
$\large\frac{dy}{dx}=\frac{2}{y}$
Step 2:
Slope of the tangent $=\large\frac{2}{y}$$=m-----(1) (x_1,y_1) lies on y^2=4x y_1^2=4x_1-----(2) Equation of tangent (x_1,y_1) y-y_1=m(x-x_1)-----(3) y=mx-mx_1+y_1 We are given the equation of tangent y=mx+1------(4) Step 3: Comparing (3) and (4) y_1-mx_1=1------(5) From (1) & (2) m=\large\frac{2}{y_1} x_1=\large\frac{y_1^2}{4} Step 4: Put these value in equation (5) y_1-\large\frac{2}{y_1}.\frac{y_1^2}{4}$$=1$
$y_1-\large\frac{y_1}{2}$$=1 \large\frac{2y_1-y_1}{2}$$=1$
$\large\frac{y_1}{2}$$=1 \Rightarrow y_1=2 m=\large\frac{2}{y_1}=\frac{2}{2}$$=1$
Part (A) is the correct answer.