Question

If the slope of one of the lines represented by $a{x}^2+2hxy+b{y}^2$ be the nth power of the other, prove that
${\left(a{b}^n\right)}^{\frac{1}{n+1}}+{\left(a^{n}b\right)}^{\frac{1}{n+1}}+2h=0$

Answer 1

This represents pair of straight lines through origin

Let $y=m_{1}x$ $y=m_{2}x$ be two lines

$\downarrow$ $\downarrow$

$mx$ $m^{n}x$

$(y-mx)(y-m^{n}x)=0$

$y^2-(m^n+m)xy+m^{n+1}{x}^2=0$

$a=1$ $b=m^{n+1}$ $h=-\frac{(m^n+m)}{2}$

$=(a{b}^n{)}^{\frac{1}{n+1}}+(a^{n}b{)}^{\frac{1}{n+1}}+2h$

$=m^n+m+\frac{2}{2}(-m^n-m)$

$=m^n-m^n+m-m$

$=0+0$

$=0$