Question

Find the measure of acute angle between the lines represented by
$\left(a^2-3b^2\right)x^2+8abxy+\left(b^2-3a^2\right)y^2=0$

Answer 1

Comparing the equation
$(a^2+3b^2)x^2+8abxy+(b^2-3a^2)y^2=0$ with
$a{x}^2-2hxy+b{y}^2=0,$ we have,
$A=a^2+3b^2,h=4ab,$ and $B=b^2-3a^2.$
$\therefore H^2-AB=16a^2{b}^2-(a^2-3b^2)(b^2-3a^2)$
$=16a^2{b}^2+(a^2-3b^2)(3a^2-b^2)$
$=16a^2{b}^2+3a^4-10a^2{b}^2+3b^4$
$=3a^4+6a^2{b}^2+3b^4$
$3(a^4+2a^2{b}^2+b^4)$
$3(a^2+b^2{)}^2$
$\therefore \sqrt{H^2-AB}=\sqrt{3}(a^2+b^2)$
Also, $A+B=(a^2-3b^2)+(b^2-3a^2)$
$=-2(a^2+b^2)$
Let $\theta$ be the acute angle between the lines.
$\therefore \tan \theta =\left|\frac{2\sqrt{H^2-AB}}{A+B}\right|$
$=\left|\frac{2\sqrt{3}(a^2+b^2)}{-2(a^2+b^2)}\right|$
$\sqrt{3}=\tan {60}^o$
$\therefore \theta ={60}^o$