Explanation

(b)
Let $y = mx + \frac{1}{m}$ is a tangent to $y^{2} = 4ax$
Equation of normal to the parabola $x^{2} = 4by$ at $(x_{1},\ y_{1})$ is
$y - y_{1} = - \frac{2b}{x_{1}}(x - x_{1})$ and $x_{1}^{2} = 4by_{1}$
$\Rightarrow y - \frac{x_{1}^{2}}{4b} = - \frac{2b}{x_{1}}(x - x_{1})$
$\Rightarrow y = - \frac{2b}{x_{1}}x + \frac{x_{1}^{2}}{4b} + 2b$
On comparing with $y = mx + \frac{1}{m},$ we get
$m = - \frac{2b}{x_{1}}$ …(i)
$\frac{x_{1}^{2}}{4b} + 2b = \frac{1}{m}$ …(ii)
From Eqs. (i) and (ii), we get
$\frac{{4b}^{2}}{m^{2}4b} + 2b = \frac{1}{m}$
$\Rightarrow b + 2bm^{2} = m$
$\Rightarrow 2bm^{2} - m + b = 0$
For real values of $m,\ D > 0$
$\Rightarrow 1 - 8b^{2} > 0 \Rightarrow b^{2} < \frac{1}{8} \Rightarrow |b| < \frac{1}{2\sqrt{2}}$