Question

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

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

Answer 1

Answer: (B)

$y^2=x$
$\Rightarrow 2y\frac{dy}{dx}=1$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2y}$
$xy=c$
$\Rightarrow x\frac{dy}{dx}+y\cdot 1=0$
$\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$
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 $=-1$)
$\Rightarrow x_1=\frac{1}{2}$
$\therefore y_1^2=x_1=\frac{1}{2}$
Since $x_1{y}_1=c$
$\therefore \frac{1}{2},\frac{1}{\sqrt{2}}=c$
$\Rightarrow c=\frac{1}{2\sqrt{2}}$.