If the curve shown below has the equation, y=(x−p)(x^3 −bx−c) then find q/p in terms of b and c.


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Question Number 98925 by ajfour last updated on 17/Jun/20

$I f t h e c u r v e s h o w n b e l o w h a s t h e$ $e q u a t i o n, y = (x - p) (x^{3} - b x - c)$ $t h e n f i n d q / p i n t e r m s o f b a n d c .$
Commented by ajfour last updated on 17/Jun/20

	Answered by mr W last updated on 17/Jun/20

	$e q n . o f t a n g e n t l i n e :$ $\frac{x}{p} + \frac{y}{q} = 1$ $\Rightarrow y = q (1 - \frac{x}{p})$ $q (1 - \frac{x}{p}) = (x - p) (x^{3} - b x - c)$ $\Rightarrow x^{3} - b x - c + \frac{q}{p} = 0$ $i t s h o u l d h a v e o n e s i n g l e a n d o n e$ $d o u b l e r o o t .$ $x^{3} - b x - c + \frac{q}{p} = (x - u) {(x - v)}^{2}$ $u + 2 v = 0$ $2 u v + v^{2} = - b$ ${u v}^{2} = c - \frac{q}{p}$ $\Rightarrow - 4 v^{2} + v^{2} = - b \Rightarrow v^{2} = \frac{b}{3}$ $\Rightarrow - 2 v^{3} = c - \frac{q}{p} \Rightarrow v^{3} = \frac{1}{2} (\frac{q}{p} - c)$ $\Rightarrow \frac{1}{4} {(\frac{q}{p} - c)}^{2} = {(\frac{b}{3})}^{2}$ $\Rightarrow \frac{q}{p} = c + \frac{2 b}{3} \sqrt{\frac{b}{3}}$
	Commented by mr W last updated on 17/Jun/20

	Commented by ajfour last updated on 17/Jun/20

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