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.

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.
