The line $y = x + 1$ is a tangent to the curve ${y^2} = 4x$ at the point

(A) $(1,\;2)$

(B) $(2,1)$

(C) $(1,\; - 2)$

(D) $( - 1,2)$

Option A is correct

We have,
${y^2} - 4x = 0$ …(i)

Slope of the line $y = x + 1$ is 1.
From (i), $2y\cfrac{{dy}}{{dx}} = 4 \Rightarrow {\rm{ }}\cfrac{{dy}}{{dx}} = \cfrac{4}{{2y}} = \cfrac{2}{y}$

$\Rightarrow$ Slope of tangent to (i) is $\cfrac{2}{y}$

$\therefore \& \cfrac{2}{y} = 1 \Rightarrow y = 2$

When $y = 2$, then from (i)${2^2} = 4x \Rightarrow x = 1$

Therefore the required point on the curve (i) is $(1,\;2)$.