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 + 2hxy

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|$.First,calculate $a + b$:$a + b = -\sin^2 \alpha + \sin^2 \alpha = 0$.Since the sum of the coefficients of $x^2$ and $y^2$ is zero $(a + b = 0)$,the lines are perpendicular to each other.Therefore,$	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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