If $4 x^{2} + p y^{2} = 45$ and $x^{2} - 4 y^{2} = 5$ cut orthogonal, then the value of p is?

1 / 9

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1 / 3

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3

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9

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Solution

The correct option is A $9$
If they cut orthogonal, then $m_{1} . m_{2} = - 1$ where $m_{1}, m_{2}$ are the slopes of the tangents to the curves.
$4 x^{2} + p y^{2} = 45$ and $x^{2} - 4 y^{2} = 5$
$20 + 16 y^{2} + p y^{2} = 45$
$y^{2} = \frac{25}{16 + p}$ and $x^{2} = \frac{180 + 5 p}{16 + p}$
$m_{1} = \frac{- 4 x}{y p}$ and $m_{2} = \frac{x}{4 y}$
$m_{1} . m_{2} = \frac{x^{2}}{y^{2} p} = 1$
$\frac{180 + 5 p}{25 p} = 1$
$p = 9$

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