The angle between the pair of straight lines | vedclass

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(D) The given equation is $x^2(\cos^2 	heta - 1) - xy \sin^2 	heta + y^2 \sin^2 	heta = 0$.Since $\cos^2 	heta - 1 = -\sin^2 	heta$,the equation

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$\frac{\pi}{2}$

Vedclass · Answer

(D) The given equation is $x^2(\cos^2 	heta - 1) - xy \sin^2 	heta + y^2 \sin^2 	heta = 0$.Since $\cos^2 	heta - 1 = -\sin^2 	heta$,the equation becomes $-x^2 \sin^2 	heta - xy \sin^2 	heta + y^2 \sin^2 	heta = 0$.Dividing by $-\sin^2 	heta$ (assuming $\sin 	heta 
eq 0$),we get $x^2 + xy - y^2 = 0$.Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 1$,$2h = 1$,and $b = -1$.The angle $\alpha$ between the lines is given by $	an \alpha = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Here,$a + b = 1 + (-1) = 0$.Since the denominator is $0$,$	an \alpha$ is undefined,which means $\alpha = \frac{\pi}{2}$.

The angle between the pair of straight lines $y^2 \sin^2 \theta - xy \sin^2 \theta + x^2(\cos^2 \theta - 1) = 0$ is

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