The condition that the curves $a{x}^2+b{y}^2=1$ and  $a_1{x}^2+b_1{y}^2=1$ may cut each other orthogonally is

The given curves are $a{x}^2+b{y}^2=1$ …………….(1)

And $a_1{x}^2+b_1{y}^2=1$ …………..(2)

From (1) $2ax+2by\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{ax}{by}=m_1$ (say)

From (2) $2a_{1}x+2b_{1}y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{a_{1}x}{b_{1}y}=m_2$ (say)

Since the curves are orthogonal , $m_1{m}_2=-1$

$\therefore \left(\frac{-ax}{by}\right)\left(-\frac{a_{1}x}{b_{1}y}\right)=-1$ $\Rightarrow a{a}_1{x}^2=-b{b}_1{y}^2$ ……………(3)

Solving (1) and (2) we get $\left(a-a_1\right)x^2=\left(b_1-b\right)y^2\mathrm{..............}\left(4\right)$

Dividing by (3), $\frac{a-a_1}{a{a}_1}=\frac{b-b_1}{b{b}_1}$