If the angle between the two lines represente | vedclass

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

Question

(A) The general equation of a pair of straight lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with $2x^2 + 5xy + 3y^2 + 6x + 7y + 4

Vedclass · Accepted Answer

$1/5$

Vedclass · Answer

(A) The general equation of a pair of straight lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$,we get $a = 2$,$2h = 5 \implies h = 5/2$,and $b = 3$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Substituting the values: $	an 	heta = \left| \frac{2\sqrt{(5/2)^2 - (2)(3)}}{2 + 3} ight|$.$	an 	heta = \left| \frac{2\sqrt{25/4 - 6}}{5} ight| = \left| \frac{2\sqrt{1/4}}{5} ight| = \left| \frac{2(1/2)}{5} ight| = \frac{1}{5}$.Since $	heta = 	an^{-1} m$,we have $m = 1/5$.

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

Similar Questions

If $\theta$ is the acute angle between the pair of lines $12x^2 + 2hxy + 7y^2 = 0$ and $\tan \theta = \frac{8}{19}$,then $h =$

The angle between the lines represented by the equation $4x^2 - 24xy + 11y^2 = 0$ is given by:

If $ax^2 + 6xy + by^2 - 10x + 10y - 6 = 0$ represents a pair of perpendicular lines,then $|a| =$

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

For $\alpha \in [0, \frac{\pi}{2}]$,the angle between the lines represented by $[x \cos \theta - y][(\cos \theta + \tan \alpha) x - (1 - \cos \theta \tan \alpha) y] = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module