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

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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

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