If the curves x^2+p y^2=1 and q x^2+y^2=1 are | vedclass

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(D) Given curves are $x^2+p y^2=1$ $(i)$ and $q x^2+y^2=1$ (ii). On differentiating $(i)$ with respect to $x$, we get $2x + 2py \frac{dy}{dx} = 0$, so

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$\frac{1}{p}+\frac{1}{q}=2$

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(D) Given curves are $x^2+p y^2=1$ $(i)$ and $q x^2+y^2=1$ (ii). On differentiating $(i)$ with respect to $x$, we get $2x + 2py \frac{dy}{dx} = 0$, so $m_1 = \frac{dy}{dx} = -\frac{x}{py}$. On differentiating (ii) with respect to $x$, we get $2qx + 2y \frac{dy}{dx} = 0$, so $m_2 = \frac{dy}{dx} = -\frac{qx}{y}$. Since the curves are orthogonal, $m_1 \cdot m_2 = -1$. Substituting the slopes: $(-\frac{x}{py}) \cdot (-\frac{qx}{y}) = -1 \Rightarrow \frac{qx^2}{py^2} = -1 \Rightarrow qx^2 = -py^2$. From $(i)$, $x^2 = 1 - py^2$. Substituting this into the orthogonality condition: $q(1 - py^2) = -py^2 \Rightarrow q - qpy^2 = -py^2 \Rightarrow q = y^2(qp - p) \Rightarrow y^2 = \frac{q}{p(q-1)}$. Similarly, $x^2 = \frac{p(1-q)}{q-p}$. Substituting $x^2$ and $y^2$ into $q x^2 + y^2 = 1$: $q(\frac{p-pq}{q-p}) + \frac{q}{pq-p} = 1$. Solving this leads to $p+q = 2pq$, which implies $\frac{1}{p} + \frac{1}{q} = 2$.

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

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