Find the angle between the lines represented | vedclass

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(A) Let the angle between the lines represented by $x^2 - 2pxy + y^2 = 0$ be $	heta$.Comparing the given equation with $ax^2 + 2hxy + by^2 = 0$,we ge

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$sec^{-1}(p)$

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(A) Let the angle between the lines represented by $x^2 - 2pxy + y^2 = 0$ be $	heta$.Comparing the given equation with $ax^2 + 2hxy + by^2 = 0$,we get $a = 1, b = 1$,and $2h = -2p$,which implies $h = -p$.The formula for the angle $	heta$ between the pair of lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Substituting the values,we get $	an 	heta = \left| \frac{2\sqrt{(-p)^2 - (1)(1)}}{1 + 1} ight| = \left| \frac{2\sqrt{p^2 - 1}}{2} ight| = \sqrt{p^2 - 1}$.Since $	an^2 	heta = p^2 - 1$,we have $p^2 = 1 + 	an^2 	heta = \sec^2 	heta$.Thus,$\sec 	heta = p$,which means $	heta = \sec^{-1}(p)$.

Find the angle between the lines represented by the equation $x^2 - 2pxy + y^2 = 0$.

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