Orthogonal Trajectory of $x^2 + 2y^2 = k^2$ $x^2 + 2y^2 = k^2$
I first take the derivative like the instructions say.
$2x + 4y \frac{dy}{dx} = 0$
I am not entirely sure why a dy and dx appears but it does in the instructions so I go with it.
Now I need to solve for $y'$
$ + 4y \frac{dy}{dx} = -2x$
$  \frac{dy}{dx} = \frac{-x}{2y}$
$  \frac{dy}{dx} = \frac{-x}{2y}$
So now I need to find the inverse negative
$  \frac{dy}{dx} = \frac{2y}{x}$
And that should be my slope at each line.
And now I need to solve that.
$  \frac{1}{2}ydy = xdx$
Take the integratal and I get
$\frac{1}{4}y^2 = \frac{1}{2}x^2 + C$
$y^2 = 2x^2 + C$
$y = \sqrt{2x^2 + C}$
This is wrong and I do not know why.

From the line David noted you; You would have $\frac{dy}{2y}=\frac{dx}{x}$. So by integrating from both sides, you get $\frac{1}{2}$Ln$|y|$=Ln$|x|+c$:
$$\int\frac{dy}{2y}=\int\frac{dx}{x}$$ $$\frac{1}{2}\int\frac{dy}{y}=\int\frac{dx}{x}$$  $$\frac{1}{2}Ln|y|=Ln|x|+c$$ wherein $c$ is a constant. You can simplify the result as $y=cx^2$.

