Solution verification: Find the orthogonal trajectories of the family of curves for $x^2 + 2y^2 = k^2$ I need help with the following question: 

Find the orthogonal trajectories of the family of curves for $x^2 + 2y^2 = k^2$

I have taken the following steps, are they correct? From what I understand, I have to take the following steps


*

*First differentiate to find the differential equation

*Then write the differential equation in this
$$\frac{dy}{dx} = -\frac{1}{f(x,y)}$$

*Then solve the new equation. 


So, this is what I have: 
The first step is to differentiate with respect to $x$ so find $\frac{dy}{dx}$. 
\begin{align}
\frac{dy}{dx} x^2 + \frac{dy}{dx} 2y^2 &= \frac{dy}{dx}k^2 \\
2x + \frac{dy}{dx}\cdot 4y &= 0 \\
\frac{dy}{dx} &= -\frac{x}{2y}
\end{align}
Now, the second step is to do the negative recipricol.
$$\frac{dy}{dx} = \frac{2y}{x}$$
The third step is to solve the newly formed differential equation. 
\begin{align}
dy \frac{1}{2y} &= \frac{1}{x} dx \\
\int \frac{1}{2y} dy &= \int \frac{1}{x} dx \\
\frac{\ln{y}}{2} &= \ln{x} + C \\
\ln y &= 2 \ln x + C\\
\ln y &= \ln x^2 + C \\
y &= Ax^2 \quad
\end{align}
In this case, $A = e^C$. Are my steps correct? 
Thanks a bunch for your help!
EDIT: 
I have a found a minor error in the last few steps when multiplying the two. It should be:
\begin{align}
\ln y &= 2(\ln x + C) \\
\ln y &= 2\ln x + 2C \\
e^{\ln y} &= e^{\ln x^2 + 2C} \\
y &= x^2 \cdot e^{2C} \\
\end{align}
Here, we let $A = e^{2C}$ and say the final answer is $$y = Ax^2$$

From the family of ellipses 
$$x^2+2y^2=k^2 ......(1)$$
you obtained the family of parabolas 
$$y=Ax^2......(2)$$
Setting $x=Y\sqrt{2}$ and $y=X/\sqrt{2}$ in (1), we obtain:
$$2Y^2+X^2=k^2 ......(3)$$
So we obtain another family of parabolas 
$$Y=B_1X^2......(4)$$
Which is equivalent to
$$x=By^2......(5)$$
This is because the original family of curves are symmetric in swapping $x$ and $y\sqrt{2}$.
