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-

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

T h a n k s S i r, p e r f e c t!

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