The angle between the lines sin^{2} alpha cdo | vedclass

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Question

(A) The given equation of the pair of straight lines is $(\cos^{2} \alpha - 1) x^{2} - 2 \cos^{2} \alpha \cdot xy + \sin^{2} \alpha y^{2} = 0$. Compar

Vedclass · Accepted Answer

$90^{\circ}$

Vedclass · Answer

(A) The given equation of the pair of straight lines is $(\cos^{2} \alpha - 1) x^{2} - 2 \cos^{2} \alpha \cdot xy + \sin^{2} \alpha y^{2} = 0$. Comparing this with the general equation $ax^{2} + 2hxy + by^{2} = 0$,we get: $a = \cos^{2} \alpha - 1 = -\sin^{2} \alpha$,$h = -\cos^{2} \alpha$,$b = \sin^{2} \alpha$. Let $	heta$ be the angle between the lines. The formula for the angle is $	an 	heta = \left| \frac{2 \sqrt{h^{2} - ab}}{a + b} ight|$. Substituting the values: $a + b = -\sin^{2} \alpha + \sin^{2} \alpha = 0$. Since the denominator is $0$,the value of $	an 	heta$ is undefined,which implies $	heta = 90^{\circ}$. Thus,the lines are perpendicular.

List-$I$	List-$II$
$(A)$ $5x^2 + 2\sqrt{7}xy - y^2 = 0$	$(I)$ $\frac{\sqrt{3}}{2}$
$(B)$ $x^2 + \sqrt{11}xy + 2y^2 = 0$	$(II)$ $\frac{1}{2\sqrt{3}}$
$(C)$ $x^2 + 2\sqrt{2}xy + y^2 = 0$	$(III)$ $\frac{1}{2}$
$(D)$ $3x^2 + 4\sqrt{2}xy + y^2 = 0$	$(IV)$ $\frac{2}{3}$
	$(V)$ $\frac{1}{\sqrt{2}}$

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

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