The equation of the family of curves which intersect the hyperbola $x y=2$ orthogonally is |
$y=\frac{x^3}{6}+C$ $y=\frac{x^2}{4}+C$ $y=\frac{-x^3}{6}+C$ $y=\frac{-x^2}{4}+C$ |
$y=\frac{x^3}{6}+C$ |
We have, $x y=2 \Rightarrow y=\frac{2}{x} \Rightarrow \frac{d y}{d x}=-\frac{2}{x^2}$ Let $y=f(x)$ be the required family of curves. Then, $\left(\frac{d y}{d x}\right)_{C_1} \times\left(\frac{d y}{d x}\right)_{C_2}=-1$ $\Rightarrow \frac{d y}{d x} \times \frac{-2}{x^2}=-1 \Rightarrow \frac{d y}{d x}=\frac{x^2}{2} \Rightarrow y=\frac{x^3}{6}+C$ This is the required family of curves. |