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)$$q x^2+y^2=1$ (ii)Differentiating $(i)$ with respect to $x$:$2x + 2py \frac{dy}{dx} = 0 \implies \frac{dy}{dx}

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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)$$q x^2+y^2=1$ (ii)Differentiating $(i)$ with respect to $x$:$2x + 2py \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = m_1 = -\frac{x}{py}$Differentiating (ii) with respect to $x$:$2qx + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = m_2 = -\frac{qx}{y}$Since the curves are orthogonal,$m_1 \cdot m_2 = -1$:$(-\frac{x}{py}) \cdot (-\frac{qx}{y}) = -1$$\frac{qx^2}{py^2} = -1 \implies qx^2 = -py^2$From $(i)$,$x^2 = 1 - py^2$. Substituting this into the condition:$q(1 - py^2) = -py^2$$q - qpy^2 = -py^2$$q = y^2(qp - p) \implies y^2 = \frac{q}{p(q-1)}$Similarly,$x^2 = \frac{p(1-q)}{q-p}$. Substituting $x^2$ and $y^2$ into $q x^2 + y^2 = 1$:$q(\frac{p(1-q)}{q-p}) + \frac{q}{p(q-1)} = 1$After simplifying,we get $p+q = 2pq$,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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