#### Answer

$y=Cx^2$

#### Work Step by Step

Given that:
$u=x^2+2y^2$
Differentiate with respect to x:
$\frac{du}{dx}=2x+4y\frac{dy}{dx}$
$0=2x+4y\frac{dy}{dx}$
so$\frac{dy}{dx}=-\frac{x}{2y}$
For orthogonal trajectory:
$\frac{dy}{dx}=\frac{-1}{f(x,y)}$
$\frac{dy}{dx}=\frac{2y}{x}$
On integrating the above term:
$\int\frac{dy}{y}=2\int\frac{dx}{x}$
$\ln{y}=2\ln{x}+c$
$y=e^(2\ln{x})\times e^c$
On simplifying above:
$y=x^2\times C$
$y=Cx^2$
Thus, the final answer: $y=Cx^2$