If the curve


If the curve ay $+x^{2}=7$ and $x^{3}=y$ cut orthogonally at $(1,1)$, then a is equal to

A. 1

B. $-6$

C. 6

D. 0


Given that the curves $a y+x^{2}=7$ and $x^{3}=y$

Differentiating both of them w.r.t. $x$,

$a \frac{d y}{d x}+2 x=0$ and $3 x^{2}=\frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{-2 x}{a}$ and $\frac{d y}{d x}=3 x^{2}$

For $(1,1)$

$\frac{d y}{d x}=\frac{-2}{a}$ and $\frac{d y}{d x}=3$

Let $\mathrm{m}_{1}=\frac{-2}{a}$ and $\mathrm{m}_{2}=3$

$m_{1} \times m_{2}=-1$

(because curves cut each other orthogonally )

$\Rightarrow \frac{-6}{a}=-1$

$\Rightarrow a=6$

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