Question:
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
Solution:
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 )
$\Rightarrow \frac{-6}{a}=-1$
$\Rightarrow a=6$