The angle between the lines represented by the equation $x^{2} - 2 p x y + y^{2} = 0$ , is

$s e c^{- 1} p$

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$c o s^{- 1} p$

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$t a n^{- 1} p$

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$t a n^{- 1} 5 p$

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Solution

The correct option is A
$s e c^{- 1} p$

$tan θ = \frac{2 \sqrt{h^{2} - a b}}{a + b}$
$tan θ = \frac{\pm 2 \sqrt{p^{2} - 1}}{1 + 1} = \pm \sqrt{p^{2} - 1} \Rightarrow θ = s e c^{- 1} p$ .

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