The angle between the lines ab(x^2 - y^2) + ( | vedclass

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

Question

(A) The given equation is $ab(x^2 - y^2) + (a^2 - b^2)xy = 0$. Expanding this,we get $abx^2 + (a^2 - b^2)xy - aby^2 = 0$. This is a homogeneous equati

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(A) The given equation is $ab(x^2 - y^2) + (a^2 - b^2)xy = 0$. Expanding this,we get $abx^2 + (a^2 - b^2)xy - aby^2 = 0$. This is a homogeneous equation of the second degree in $x$ and $y$ of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = ab$,$2H = a^2 - b^2$,and $B = -ab$. The condition for the lines to be perpendicular is $A + B = 0$. Here,$A + B = ab + (-ab) = 0$. Since the sum of the coefficients of $x^2$ and $y^2$ is zero,the lines represented by the equation are perpendicular to each other. Therefore,the angle between the lines is $\frac{\pi}{2}$.

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

Similar Questions

If the angle $2 \theta$ is acute,then the acute angle between the pair of straight lines $x^2(\cos \theta - \sin \theta) + 2xy \cos \theta + y^2(\cos \theta + \sin \theta) = 0$ is:

If $\theta$ is an acute angle between the lines $k x^2 - 4 x y + y^2 = 0$ and $\tan \theta = \frac{1}{2}$,then the value of $k$ is

If the slopes of both the lines given by $x^2 + 2hxy + 6y^2 = 0$ are positive and the angle between these lines is $\operatorname{Tan}^{-1}\left(\frac{1}{7}\right)$,then find the value of $h$.

If the acute angle between the lines $ax^2+2hxy+by^2=0$ is $\frac{\pi}{4}$,then $4h^2=$

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