Browse Questions

# Choose the correct answer in the line $y = x + 1$ is a tangent to the curve $y^2 = 4x$ at the point

$(A)\; (1, \: 2) \quad (B) \;(2, \: 1) \quad (C) \;(1, \: -2) \quad (D)\; (-1, \: 2)$

Toolbox:
• If a tangent line to the curve $y=f(x)$ makes an angle $\theta$ with $x$-axis in the +ve direction,then $\large\frac{dy}{dx}$=slope of the tangent=$\tan \theta$
• Slope of the normal is $\large\frac{-1}{\Large\frac{dy}{dx}}$
Step 1:
Equation of the curve is $y^2=4x$
Differentiating w.r.t $x$ we get,
$2y\large\frac{dy}{dx}$$=4 Therefore \large\frac{dy}{dx}=\frac{4}{2y}=\frac{2}{y} \Rightarrow m=\large\frac{2}{y} Step 2: Equation of the line is y=x+1----(1) Slope of the given line is 1. \Rightarrow \large\frac{2}{y}$$=1$
Therefore $y=2$
Step 3:
When $y=2$,substituting for $y$ in equ(1)
$2=x+1$
$\Rightarrow x=1$
Hence the required point is $(1,2)$
The correct option is $A$.