Explanation

(b)
On differentiating given curves respectively
$2x = - 9A\frac{dy}{dx}\ \ \ \Rightarrow \ \ \ \ \ \ \ \ \left( \frac{dy}{dx} \right)_{c_{1}} = - \frac{2x}{9A}$
And $2x = A\ \frac{dy}{dx}$
$\Rightarrow \ \ \left( \frac{dy}{dx} \right)_{c_{2}} = \frac{2x}{A}$
For orthogonally,
$\left( \frac{dy}{dx} \right)_{c_{1}}\ \left( \frac{dy}{dx} \right)_{c_{2}} = - 1$
$\Rightarrow \ \ \left( - \frac{2x}{9A} \right)\left( \frac{2x}{A} \right) = - 1$ ...(i)
From given curves,
$9A(9 - y) = A(y + 1)\ \ \ \ \ \Rightarrow \ \ \ y = 8$
Since, $x^{2} = 9A(9 - y)$
$\Rightarrow \ \ \ \ \ \ x^{2} = 9A(9 - 8) = 9A$
From Eq. (i), $\frac{4x^{2}}{9A^{2}} = 1$
$\Rightarrow \ \ \ \ \ \ \frac{4.9\ A}{9A^{2}} = 1\ \ \ \ \ \Rightarrow \ \ \ \ \ A\text{=}4$