The angle between the lines represented by (s | vedclass

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Question

(C) The given equation is $(\sin ^2 \alpha) y^2 - 2xy(\cos ^2 \alpha) + (\cos ^2 \alpha - 1) x^2 = 0$.Comparing this with the standard form $ax^2 + 2h

Vedclass · Accepted Answer

$90^{\circ}$

Vedclass · Answer

(C) The given equation is $(\sin ^2 \alpha) y^2 - 2xy(\cos ^2 \alpha) + (\cos ^2 \alpha - 1) x^2 = 0$.Comparing this with the standard form $ax^2 + 2hxy + by^2 = 0$,we get:$a = \cos ^2 \alpha - 1 = -\sin ^2 \alpha$$h = -\cos ^2 \alpha$$b = \sin ^2 \alpha$The angle $	heta$ between the pair of lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Calculating the denominator: $a + b = -\sin ^2 \alpha + \sin ^2 \alpha = 0$.Since the denominator is $0$,the lines are perpendicular to each other.Therefore,$	an 	heta = \infty$,which implies $	heta = 90^{\circ}$.

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

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