Explanation

(b)
Given, $y^{2} = x^{3}$
On differentiating w.r.t $x$, we get
$2y\frac{dy}{dx} = 3x^{2}$
$\Rightarrow \ \ \ \ \frac{dy}{dx} = \frac{3x^{2}}{2y}\ \ \ \Rightarrow \ \ \ \ \ \left( \frac{dy}{dx} \right)_{(1,\ 1)} = \frac{3}{2}$
$\therefore$ Equation of normal at point (1, 1) is
$y - 1 = - \frac{2}{3}(x - 1)$
$\Rightarrow \ \ \ \ 2x + 3y = 5$
Hence, the equation of line parallel to above line will be
in option (b), $ie,\ \ \ \ 2x + 3y = 7$