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

Correct answer: (A) (π)/2. We have, a b (x^2-y^2 )+ (a^2-b^2 ) x y=0 a b x^2+ (a^2-b^2 ) x y-a b y^2=0 Sum of the coefficient of x^2 and y^2 i.e. a b-a b=0 ∴ Lines are perpendicular to each other. ∴ Angle between lines =π / 2

Q89 / 160 · AP EAMCET 2021 (25 Aug Shift 1) · Mathematics

The angle between the lines $a b\left(x^2-y^2\right)+\left(a^2-b^2\right) x y=0$ is

Correct answer
A
$\frac{\pi}{2}$
B
$\frac{\pi}{3}$
C
$\frac{\pi}{4}$
D
$\frac{\pi}{6}$

Solution

We have, $$ \begin{aligned} & a b\left(x^2-y^2\right)+\left(a^2-b^2\right) x y=0 \\ & a b x^2+\left(a^2-b^2\right) x y-a b y^2=0 \end{aligned} $$ Sum of the coefficient of $x^2$ and $y^2$ $$ \text { i.e. } a b-a b=0 $$ $\therefore \quad$ Lines are perpendicular to each other. $\therefore \quad$ Angle between lines $=\pi / 2$

Hence, the correct answer is option: 'A'.

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