Explanation

(d)
$\because$ Equation of normal at $(x,y)$ is
$Y - y = \frac{dx}{dy}(X - x)$
Put, $y = 0$
Then, $X = x + y\ \frac{dy}{dx}$
Given, $y^{2} = 2x\ X$
$\Rightarrow y^{2} = 2x\left( \ x + y\ \frac{dy}{dx}\ \right)$
$\Rightarrow \ \frac{dy}{dx} = \frac{y^{2} - 2x^{2}}{2xy} = \frac{\left( \frac{y}{x} \right)^{2} - 2}{2\left( \frac{y}{x} \right)}$
Put $y = vx$, we get
$\frac{dy}{dx} = v + x\frac{dv}{dx}$
Then, $v + x\frac{dv}{dx} = \frac{v^{2} - 2}{2v}$
$\ \Rightarrow x\frac{dv}{dx} = - \frac{(2 + v^{2})}{2v}$
$\Rightarrow \ \frac{2v\ dv}{(2 + v^{2})} + \frac{dv}{x} = 0$
On integrating both sides, we get
In $\left( 2 + v^{2} \right) + In|x| = In\ c$
$\Rightarrow In\ \left( |x|\left( 2 + v^{2} \right) \right) = In\ c$
$\Rightarrow |x|\left( \ 2 + \frac{y^{2}}{x^{2}} \right) = c$
$\because$ It passes through (2, 1), then
$2\left( 2 + \frac{1}{4} \right) = c$
$\Rightarrow c = \frac{9}{2}$
Then, $|x|\left( \ 2 + \frac{y^{2}}{x^{2}} \right) = \frac{9}{2}$
$\Rightarrow 2x^{2} + y^{2} = \frac{9}{2}|x|$
$\Rightarrow 4x^{2} + 2y^{2} = 9|x|$