#### If the curve $ay + x^2 = 7$ and $x^3 = y$, cut orthogonally at (1, 1), then the value of a is: A. 1 B. 0 C. – 6 D. 6

Given the fact that curve $ay + x^2 = 7$ and $x^3 = y$, cut orthogonally at (1, 1)

Differentiate on both sides with x and get

$\frac{\mathrm{d}\left(\mathrm{ay}+\mathrm{x}^{2}\right)}{\mathrm{dx}}=\frac{\mathrm{d}(7)}{\mathrm{dx}}$

Apply sum rule and also 0 is the derivative of the constant, so
$\Rightarrow \frac{\mathrm{d}(\mathrm{ay})}{\mathrm{dx}}+\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}=0$
Apply power rule and get
$\\\Rightarrow \mathrm{a} \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{x}=0\\\Rightarrow \frac{d y}{d x}=-\frac{2 x}{a}$
Putting (1,1)

$\\\Rightarrow\left(\frac{d y}{d x}\right)_{(1,1)}=-\frac{2 x}{a} \\\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,1)}=-\frac{2(1)}{\mathrm{a}}=-\frac{2}{\mathrm{a}} \ldots \ldots(\mathrm{i}) \\x^{3}=y$
Differentiate on both sides with x and get
$\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}=\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dx}}$

Apply power rule and get
$\Rightarrow 3 \mathrm{x}^{2}=\frac{\mathrm{dy}}{\mathrm{dx}}$
Putting (1,1)
$\\\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,1)}=3 \mathrm{x}^{2} \\\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,1)}=3(1)^{2}=3 \ldots \ldots(ii)$

Both curves cut orthogonally at (1,1), 50
$\left(\frac{d y}{d x}\right)_{(1,1)} \times\left(\frac{d y}{d x}\right)_{(1,1)}=-1$
So from (i) and (ii), we get
$\\\left(-\frac{2}{a}\right) \times 3=-1 \\\Rightarrow-\frac{6}{a}=-1\\\Rightarrow a=6$

Hence when the curves cut orthogonally at (1, 1), then the value of a is 6.

So the correct answer is option D.