If theta is the acute angle between the lines | vedclass

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

Question

(C) The given equation is $x^2 - 3xy + 2y^2 = 0$. Comparing this with $ax^2 + 2hxy + by^2 = 0$, we get $a = 1$, $2h = -3$ (so $h = -\frac{3}{2}$), and

Vedclass · Accepted Answer

$-3$

Vedclass · Answer

(C) The given equation is $x^2 - 3xy + 2y^2 = 0$. Comparing this with $ax^2 + 2hxy + by^2 = 0$, we get $a = 1$, $2h = -3$ (so $h = -\frac{3}{2}$), and $b = 2$.The formula for the acute angle $	heta$ between the lines is $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Substituting the values: $	an 	heta = \left| \frac{2\sqrt{(-\frac{3}{2})^2 - (1)(2)}}{1 + 2} ight| = \left| \frac{2\sqrt{\frac{9}{4} - 2}}{3} ight| = \left| \frac{2\sqrt{\frac{1}{4}}}{3} ight| = \left| \frac{2 \cdot \frac{1}{2}}{3} ight| = \frac{1}{3}$.Since $	an 	heta = \frac{1}{3}$, we have $\sin 	heta = 1$ and $\cos 	heta = 3$ (in ratio terms) or simply divide the expression by $\cos 	heta$: $\frac{3 	an 	heta + 2}{3 	an 	heta - 2}$.Substituting $	an 	heta = \frac{1}{3}$: $\frac{3(\frac{1}{3}) + 2}{3(\frac{1}{3}) - 2} = \frac{1 + 2}{1 - 2} = \frac{3}{-1} = -3$.

If $\theta$ is the acute angle between the lines represented by the equation $x^2 - 3xy + 2y^2 = 0$, then $\frac{3 \sin \theta + 2 \cos \theta}{3 \sin \theta - 2 \cos \theta} = $

Similar Questions

If $4ab = 3h^2$, then the ratio of the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is...

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 sine of the angle between the pair of lines represented by the equation $x^2-7xy+12y^2=0$ is

The angle between the pair of lines represented by $2x^2 - 7xy + 3y^2 = 0$ is .....$^o$.

If the lines represented by $(k^2+2) x^2+3 xy-6 y^2=0$ are perpendicular to each other,then the values of $k$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module