Question

If the curves $y^2 = x$ and $xy = c $ are orthogonal, then $c = ..... (x, y \in R^+) ; ( c \neq 0)$.

• A. $-\frac{1}{2\sqrt2}$
• B. $\frac{1}{2\sqrt2}$
• C. $\frac{1}{8}$
• D. $\pm\frac{1}{2}$

Answer 1

Answer: (B)

$y^{2}=x$
$\Rightarrow 2y \frac{dy}{dx}=1$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2y}$
$xy=c$
$\Rightarrow x \frac{dy}{dx}+y\cdot1=0$
$\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$
At. $\left(x_{1}, y_{1}\right)$, product of slopes
$\frac{1}{2y_{1}}\cdot\frac{y_{1}}{x_{1}}$
$=-\frac{1}{2x_{1}}=-1$
($\because$ curves cut orthogonally $\therefore $ product of slopes $= - 1$)
$\Rightarrow x_{1}=\frac{1}{2}$
$\therefore y^{2}_{1}=x_{1}=\frac{1}{2}$
Since $x_{1}y_{1}=c$
$\therefore \frac{1}{2}, \frac{1}{\sqrt{2}}=c$
$\Rightarrow c=\frac{1}{2\sqrt{2}}$.