Show that the condition for the orthogonality of the curves $a{x}^2+b{y}^2=1$ and $a_1{x}^2+b_1{y}^2=1$ is $\frac{1}{a}-\frac{1}{b}=\frac{1}{a_1}-\frac{1}{b_1}$

The given curves are

$a{x}^2+b{y}^2=1......\left(1\right)$

$a_1{x}^2+b_1{y}^2=1.......\left(2\right)$

let the curves (1) and (2) intersect at $P\left(x_1,y_1\right).$

Then $a{x}_1^2+b{y}_1^2=1,a_1{x}_1^2+b_1{y}_1^2=1$

$\Rightarrow \left(a-a_1\right){x_1}^2+\left(b-b_1\right)y_1^2=0$

$\Rightarrow \frac{x_1^2}{y_1^2}=-\left(\frac{b-b_1}{a-a_1}\right)......\left(3\right)$

Diff (1) w.r.t. $x$ $\frac{dy}{dx}=-\frac{ax}{by}......\left(4\right)$

Diff (2) w.r.t $\mathrm{x},\frac{dy}{dx}=-\frac{a_{1}x}{b_{1}y}......\left(5\right)$

From (4),  $m_1={\left(\frac{dy}{dx}\right)}_P=-\frac{a{x}_1}{b{y}_1}$

From (5), $m_2={\left(\frac{dy}{dx}\right)}_P=-\frac{a_1{x}_1}{b_1{y}_1}$

The curves (1) and (2) are orthogonal at $P$ $\Rightarrow m_1{m}_2=-1$

$\Rightarrow \left(\frac{-a{x}_1}{b{y}_1}\right)\left(\frac{-a_1{x}_1}{b_1{y}_1}\right)=-1\Rightarrow \frac{a{a}_1}{b{b}_1}\frac{x_1^2}{y_1^2}=-1.......\left(6\right)$

From (3) and (6) we have $\frac{a{a}_1}{b{b}_1}\left(\frac{b-b_1}{a-a_1}\right)=1$

$\Rightarrow \frac{a-a_1}{a{a}_1}=\frac{b-b_1}{b{b}_1}\Rightarrow \frac{1}{a_1}-\frac{1}{a}=\frac{1}{b_1}-\frac{1}{b}$

$\Rightarrow \frac{1}{a}-\frac{1}{b}=\frac{1}{a_1}-\frac{1}{b_1}$

This is the required condition.