The acute angle between the lines (x^2+y^2) s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given equation is $(x^2+y^2) \sin 	heta + 2xy = 0$,which can be written as $(\sin 	heta)x^2 + 2xy + (\sin 	heta)y^2 = 0$. Comparing this wi

Vedclass · Accepted Answer

$\frac{\pi}{2}-	heta$

Vedclass · Answer

(C) The given equation is $(x^2+y^2) \sin 	heta + 2xy = 0$,which can be written as $(\sin 	heta)x^2 + 2xy + (\sin 	heta)y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = \sin 	heta$,$h = 1$,and $b = \sin 	heta$. Let $\alpha$ be the acute angle between the lines. The formula for the angle is $	an \alpha = \left| \frac{2\sqrt{h^2 - ab}}{a+b} ight|$. Substituting the values,we get $	an \alpha = \left| \frac{2\sqrt{1^2 - (\sin 	heta)(\sin 	heta)}}{\sin 	heta + \sin 	heta} ight|$. $	an \alpha = \left| \frac{2\sqrt{1 - \sin^2 	heta}}{2 \sin 	heta} ight| = \left| \frac{2\cos 	heta}{2 \sin 	heta} ight| = |\cot 	heta|$. Since $\alpha$ is the acute angle,$	an \alpha = 	an(\frac{\pi}{2} - 	heta)$. Therefore,$\alpha = \frac{\pi}{2} - 	heta$.

The acute angle between the lines $(x^2+y^2) \sin \theta+2xy=0$ is

Similar Questions

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

If the slope of one of the lines represented by $2x^2 + 3xy + ky^2 = 0$ is $2$,then the angle between the pair of lines is

If $\theta_1$ and $\theta_2$ are the acute angles between the pairs of lines $3x^2 - 7xy + 4y^2 = 0$ and $6x^2 - 5xy + y^2 = 0$ respectively,then:

If the lines given by $(x^2+y^2) \sin^2 \alpha = (x \cos \alpha - y \sin \alpha)^2$ are perpendicular to each other,then $\sin^2 \alpha + \tan^2 \alpha = $

Find the angle between the pair of lines represented by the equation $x^2+4xy+y^2=0$. (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module