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
$$
\begin{array}{cc}
& -\frac{p q}{p q-q}+\frac{q}{p q=p}=1 \\
\Rightarrow & -p q+q=p q-p \Rightarrow p+q=2 p q \\
\Rightarrow & \frac{1}{p}+\frac{1}{q}=2
\end{array}
$$