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.

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.

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.