(a) The equation of the curve is $\mathrm{y}^2=4 \mathrm{x}$,
Differentiating w.r.t. $x \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2 \mathrm{y}}=\frac{2}{\mathrm{y}}$,
Slope of tangent $=\frac{2}{y}=m, \therefore \quad y=\frac{2}{m} \quad$...(i)
$\left(\mathrm{x}_1, \mathrm{y}_1\right)$ lies on $\mathrm{y}^2=4 \mathrm{x}, \mathrm{y}_1{ }^2=4 \mathrm{x}_1 \quad \ldots(ii)$
Equation of tangent at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$
or
$\begin{array}{ll}y & =m x+y_1-m x_1 \quad \ldots(iii)\\ y & =m x+1 \quad \ldots (iv)\end{array}$
Comparing (iii) \& (iv) $\mathrm{y}_1-\mathrm{mx}_1=1 \quad \ldots(v)$
from (i) \&(ii) $m=\frac{2}{y_1}, x_1=\frac{y_1^2}{4}$
$\therefore \quad$ Put these values in (v) or $\mathrm{y}_1-\frac{\mathrm{y}_1}{2}=\frac{\mathrm{y}_1}{2}=1$
$\therefore \quad \mathrm{y}_1=2 \quad \therefore \mathrm{m}=\frac{2}{\mathrm{y}_1}=\frac{2}{2}=1$