If the curves $x^{2} + p y^{2} = 1 a n d q x^{2} + y^{2} = 1$ are orthogonal to each other then

$p - q = 2$

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

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

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

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Solution

The correct option is D
$\frac{1}{p} + \frac{1}{q} = 2$

Solving the two equations

$\frac{x^{2}}{y^{2}} = \frac{p - 1}{q - 1} - - - - - - - (1)$
$m_{1} . m_{2} = - 1 \Rightarrow \frac{x^{2}}{y^{2}} . \frac{q}{p} = - 1$
$\Rightarrow \frac{1}{p} + \frac{1}{q} = 2$

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Similar questions

x^{2} + p y^{2} = 1

q x^{2} + y^{2} = 1

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