The angle between the lines represented by th | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) The given equation is $y^{2} \sin^{2} 	heta - xy \sin^{2} 	heta + x^{2}(\cos^{2} 	heta - 1) = 0$. Since $\cos^{2} 	heta - 1 = -\sin^{2} 	heta$,the equation becomes $y^{2} \sin^{2} 	heta - xy \sin^{2} 	heta - x^{2} \sin^{2} 	heta = 0$. Dividing by $\sin^{2} 	heta$ (assuming $\sin^{2} 	heta 
eq 0$),we get $y^{2} - xy - x^{2} = 0$. For a general second-degree equation $ax^{2} + 2hxy + by^{2} = 0$,the angle $\phi$ between the lines is given by $	an \phi = \left| \frac{2\sqrt{h^{2} - ab}}{a + b} ight|$. Here,$a = -1$,$2h = -1$ (so $h = -1/2$),and $b = 1$. Since $a + b = -1 + 1 = 0$,the sum of the coefficients of $x^{2}$ and $y^{2}$ is zero. This implies that the lines are perpendicular to each other. Therefore,the angle between the lines is $\frac{\pi}{2}$.

The angle between the lines represented by the equation $y^{2} \sin^{2} \theta - xy \sin^{2} \theta + x^{2}(\cos^{2} \theta - 1) = 0$ is

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