The line $y = mx + 1$ is a tangent to the curve ${y^2} = 4x$ if the value of $m$ is

(A) $1$

(B) $2$

(C) $3$

(D) $112$

Option A is correct

We have, ${y^2} = 4x$ …(i) Differentiating (i) w.r.t. $x$, we get
$2y\cfrac{{dy}}{{dx}} = 4 = \cfrac{{dy}}{{dx}} = \cfrac{2}{y}$ Now,

${\left( {\cfrac{{dy}}{{dx}}} \right)_{{x_1} - {y_1}}} = \cfrac{2}{{{y_1}}} = m(say)$ …(ii)

Let $({x_1},{y_1})$ be any point on (i), then,${y_1}^2 = 4{x_1}$ …(iii)

Therefore, The equation of tangent is
$y - {y_1} = m(x - {x_1}) \Rightarrow y = mx + {y_1} - m{x_1}$ …(iv)

We have equation of tangent to curve ${y^2} = 4x$ as $y = mx + 1$
$\Rightarrow mx + 1 = mx + {y_1} - m{x_1} \Rightarrow {y_1} - m{x_1} = 1$ …(v)

Also, $m = \cfrac{2}{{{y_1}}}$ , ${x_1} = \cfrac{{y_1^2}}{4}$ [By (ii) and (iii)]
Putting these values in (v) we get,

${y_1} - \cfrac{2}{{{y_1}}} \times \cfrac{{y_1^2}}{4} = 1 = {y_1} - \cfrac{{{y_1}}}{2} = 1 = \cfrac{{{y_1}}}{2} = 1 \Rightarrow {y_1} = 2$

Therefore, $m = \cfrac{2}{{{y_1}}} = \cfrac{2}{2} = 1$