If the pair of lines given by (x^2+y^2) cos^2 | vedclass

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

Question

(B) The given equation of the pair of lines is: $(x^2+y^2) \cos^2 	heta = (x \cos 	heta + y \sin 	heta)^2$ Expanding the right side: $(x^2+y^2) \co

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) The given equation of the pair of lines is: $(x^2+y^2) \cos^2 	heta = (x \cos 	heta + y \sin 	heta)^2$ Expanding the right side: $(x^2+y^2) \cos^2 	heta = x^2 \cos^2 	heta + y^2 \sin^2 	heta + 2xy \sin 	heta \cos 	heta$ Subtracting $x^2 \cos^2 	heta$ from both sides: $y^2 \cos^2 	heta = y^2 \sin^2 	heta + 2xy \sin 	heta \cos 	heta$ Rearranging into the general form $Ax^2 + 2Hxy + By^2 = 0$: $0x^2 + (2 \sin 	heta \cos 	heta)xy + (\sin^2 	heta - \cos^2 	heta)y^2 = 0$ For the pair of lines to be perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $A + B = 0$ $0 + (\sin^2 	heta - \cos^2 	heta) = 0$ $\sin^2 	heta = \cos^2 	heta$ $	an^2 	heta = 1$ $	an 	heta = \pm 1$ Thus,$	heta = \frac{\pi}{4}$ or $\frac{3\pi}{4}$.

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

Similar Questions

The equation $12x^{2}+7xy+ay^{2}+13x-y+3=0$ represents a pair of perpendicular lines. Then the value of $a$ is

If $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of parallel lines,then $\sqrt{\frac{g^2-ac}{f^2-bc}}$ is equal to

The equation $x^2-3xy+2y^2+3x-5y+2=0$ represents a pair of straight lines. If $\theta$ is the angle between them,then the value of $\cos \theta$ is equal to

If the lines represented by $(p - q) x^2 + 2 (p + q) xy + (q - p) y^2 = 0$ are perpendicular to each other,then:

The angle between the lines $x^2 + 4xy + y^2 = 0$ is ..... $^o$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module