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

\frac{1}{9}

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9

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3

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18

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Solution

The correct option is B $9$
Equation of curves are $4 x^{2} + p y^{2} = 45$ and $x^{2} - 4 y^{2} = 5$

Let $(a, b)$ be the point of contact

Now,
$m_{1} = (\frac{d y}{d x}) = (- \frac{4 x}{p y})$

and $m_{2} = (\frac{d y}{d x}) = (\frac{x}{4 y})$

Since, both curves cut orthogonally, then

$- \frac{4 x}{p y} . \frac{x}{4 y} = - 1$

$- \frac{x^{2}}{p y^{2}} = - 1$

$x^{2} = p y^{2}$

$∴ F r o m 1 s t c u r v e 5 x^{2} = 45$

$∴ x = 3$

$∴ y = 1$

$∴ p (1) = (3)^{2} = 9 \Rightarrow p = 9.$

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