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

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Solution

We have,
$a x^{2} + 2 h x y + b y^{2} = 0$

$E q u a t i o n o f l i n e s b e y = m x \to (1) y = m^{n} x \to (2)$

$N o c o n s t a n t t e r m u s t h e y p a i r t h r o u g h o r i g i n (y - m x) (y - m^{n} x) = 0 y^{2} - (m + m^{n}) x + m^{n + 1} x^{2} = 0 \to (B)$

$C o m p a r e e q u a t i o n (A) a n d (B) a = m^{n + 1} b = 12 h = - (m^{n} + m)$

$L . H . S = {(a b^{n})}^{1 / n + 1} + {(a^{n} b)}^{1 / n + 1} + 2 h = {(m^{n + 1} {(4)}^{n})}^{1 / n + 1} + {(m^{n (n + 1)} 1)}^{1 / n + 1} - (m^{n} + m) \Rightarrow m + m^{n} - m^{n} - m$
$= 0$
$R . H . S .$

$H e n c e, p r o v e d .$

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