The angle between the pair of straight lines | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) The given equation is $x^2(\cos^2 	heta - 1) - xy \sin^2 	heta + y^2 \sin^2 	heta = 0$.Since $\cos^2 	heta - 1 = -\sin^2 	heta$,the equation becomes $-x^2 \sin^2 	heta - xy \sin^2 	heta + y^2 \sin^2 	heta = 0$.Dividing by $-\sin^2 	heta$ (assuming $\sin 	heta 
eq 0$),we get $x^2 + xy - y^2 = 0$.Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 1$,$2h = 1$,and $b = -1$.The angle $\alpha$ between the lines is given by $	an \alpha = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Here,$a + b = 1 + (-1) = 0$.Since the denominator is $0$,$	an \alpha$ is undefined,which means $\alpha = \frac{\pi}{2}$.

The angle between the pair of straight lines $y^2 \sin^2 \theta - xy \sin^2 \theta + x^2(\cos^2 \theta - 1) = 0$ is

Similar Questions

If the angle between the two lines represented by $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$ is $\tan^{-1} m$,then $m = $

If $\varphi$ is the angle between the lines $a x^{2}+2 h x y+b y^{2}=0$,then the angle between the lines $x^{2}+2 x y \sec \theta+y^{2}=0$ is

The angle between the lines $ab(x^2 - y^2) + (a^2 - b^2)xy = 0$ is

If the angle between the pair of lines given by the equation $ax^2+4xy+2y^2=0$ is $45^{\circ}$,then the possible values of $a$ are:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module