Prove that the family of curves $\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1$,where $\lambda$ is a parameter,is self orthogonal. Prove that the family of curves $\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1$,where $\lambda$ is a parameter,is self orthogonal.
I tried to find the differential equation of first order(because there is one parametr $\lambda$) corresponding to $\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1$ and then replace $\frac{dy}{dx}$ by $\frac{-dx}{dy}$ and then re-integrating it but the calculations has gone messy.
 A: Let $$\varphi_{\lambda}(x,y) = \frac{x^2}{a^2+\lambda} + \frac{y^2}{b^2+\lambda}$$ and $(u,v)$ be a point lying on the intersection of the curves
$\varphi_{\lambda_1}(x,y) = 1$ and $\varphi_{\lambda_2}(x,y) = 1$.
The normal for the $i^{th}$ curve at $(u,v)$ is in the direction
$$\left.\nabla \varphi_{\lambda_i}(x,y)\right|_{(u,v)} = \left(\frac{2u}{a^2+\lambda_i}, \frac{2v}{b^2 + \lambda_i}\right)$$ 
The dot product of these two normal vectors is proportional to 
$$\bigg[ \nabla\varphi_{\lambda_1}(x,y)\cdot\nabla\varphi_{\lambda_2}(x,y)\bigg]_{(u,v)}
= \frac{4u^2}{(a^2+\lambda_1)(a^2+\lambda_2)} + \frac{4v^2}{(b^2+\lambda_1)(b^2+\lambda_2)}\\ 
= \frac{4}{\lambda_2-\lambda_1}\left[
\left(\frac{u^2}{a^2+\lambda_1} + \frac{v^2}{b^2+\lambda_1}\right)
- \left(\frac{u^2}{a^2+\lambda_2} + \frac{v^2}{b^2+\lambda_2}\right)\right]\\
= \frac{4}{\lambda_2-\lambda_1}(1 - 1) = 0$$
As a result, the dot product itself vanishes and the normal directions for the two curves are orthogonal to each other.
A: Let us find a differential equation not involving $\lambda$ that describes the situation. Then we show that it is invariant under the change of $y'\mapsto -\frac{1}{y'}$.
Differentiating,
$$
\frac{2x}{a^2+\lambda}+\frac{2yy'}{b^2+\lambda}=0.
$$
Then, using the equation and the differentiated one, we find that
$$
a^2+\lambda=\frac{x^2y'-xy}{y'},\quad\text{and}\quad b^2+\lambda=y^2-xyy'.
$$
We eliminate $\lambda$ and find that
$$
a^2-b^2=\frac{x^2y'-xy}{y'}-y^2+xyy',
$$
which implies that the general differential equation is
$$
xy(y')^2+(x^2-y^2-a^2+b^2)y'-xy=0.
$$
If we substitute $y'$ by $-1/y'$, we get
$$
xy(-1/y')^2+(x^2-y^2-a^2+b^2)(-1/y')-xy=0,
$$
which after multiplication with $-(y')^2$ gives back
$$
-xy+(x^2-y^2-a^2+b^2)y'+xy(y')^2=0.
$$
The self-orthogonality follows.
