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

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- 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 $A$.

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