The acute angle between the lines represented | vedclass

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Question

(A) The given equation is $\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 = 0$,we get $a = \sqrt{3}

Vedclass · Accepted Answer

$\pi /6$

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(A) The given equation is $\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 = 0$,we get $a = \sqrt{3}$,$2h = -4$ (so $h = -2$),and $b = \sqrt{3}$.The formula for the angle $	heta$ between the pair of lines is $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Substituting the values: $	an 	heta = \left| \frac{2\sqrt{(-2)^2 - (\sqrt{3})(\sqrt{3})}}{\sqrt{3} + \sqrt{3}} ight|$.$	an 	heta = \left| \frac{2\sqrt{4 - 3}}{2\sqrt{3}} ight| = \frac{2(1)}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$.Since $	an 	heta = \frac{1}{\sqrt{3}}$,the acute angle is $	heta = 30^\circ$ or $\frac{\pi}{6}$ radians.

The acute angle between the lines represented by $({x^2} + {y^2})\sqrt{3} = 4xy$ is

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