If the curves $x^2+p y^2=1$ and $q x^2+y^2=1$ are orthogonal to each other, then

Given curves are $$ \begin{aligned} & x^2+p y^2=1 \\ & q x^2+y^2=1 \end{aligned} $$ and On differentiating Eq. (i), w.r.t., $x$ we get $$ \begin{gathered} 2 x+2 y p \frac{d y}{d x}=0 \\ \Rightarrow \quad \frac{d y}{d x}=m_1=-\frac{x}{p y} \end{gathered} $$ On differentiating Eq. (ii), w.r.t. to $x$, we get $$ \begin{array}{r} 2 q x+2 y \frac{d y}{d x}-0 \\ \Rightarrow \quad \frac{d y}{d x}=m_2=\frac{-q x}{y} \end{array} $$ Since, both the curves are orthogonal to each other, Then, $$ \Rightarrow \quad \frac{-x}{p y} \cdot \frac{-q x}{y}=-1 $$ $\Rightarrow \quad q x^2=-p y^2$ $\Rightarrow \quad q\left(1-p y^2\right)=-p y^2$ [from Eq. (i)] $\Rightarrow \quad q-p q y^2=-p y^2$ $\Rightarrow \quad q=(p q-p)=y^2$ $\therefore \quad y^2=\frac{q}{p q-p}$ and $$ x^2=\frac{-p}{p q-p} $$ On putting $x^2$ and $y^2$ in Eq. (ii) we get $$ \begin{array}{cc} & -\frac{p q}{p q-q}+\frac{q}{p q=p}=1 \\ \Rightarrow & -p q+q=p q-p \Rightarrow p+q=2 p q \\ \Rightarrow & \frac{1}{p}+\frac{1}{q}=2 \end{array} $$