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Question Number 63178 by ajfour last updated on 30/Jun/19

Commented by Prithwish sen last updated on 30/Jun/19

$Sir, whether the answer is [(\frac{a + b}{2}), {(\frac{a + b}{2})}^{2}] or not ?$
	Answered by mr W last updated on 01/Jul/19

	$t a n g e n t a t P :$ $m_{T} = 2 x_{P}$ $n o r m a l a t P :$ $m_{N P} = - \frac{1}{m_{T}} = - \frac{1}{2 x_{P}}$ $m_{A P} = \frac{x_{P}^{2}}{x_{P} - a}$ $m_{B P} = \frac{x_{P}^{2}}{x_{P} - b}$ $2 θ_{N P} = (θ_{A P} + θ_{B P})$ $\frac{2 m_{N P}}{1 - m_{N P}^{2}} = \frac{m_{A P} + m_{B P}}{1 - m_{A P} m_{B P}}$ $\Rightarrow \frac{- \frac{2}{2 x_{P}}}{1 - {(- \frac{1}{2 x_{P}})}^{2}} = \frac{\frac{x_{P}^{2}}{x_{P} - a} + \frac{x_{P}^{2}}{x_{P} - b}}{1 - \frac{x_{P}^{2}}{x_{P} - a} \times \frac{x_{P}^{2}}{x_{P} - b}}$ $\Rightarrow - \frac{4 x_{P}}{4 x_{P}^{2} - 1} = \frac{(2 x_{P} - a - b) x_{P}^{2}}{(x_{P} - a) (x_{P} - b) - x_{P}^{4}}$ $\Rightarrow - \frac{4}{4 x_{P}^{2} - 1} = \frac{[2 x_{P} - (a + b)] x_{P}}{x_{P}^{2} + a b - (a + b) x_{P} - x_{P}^{4}}$ $\Rightarrow - 4 a b + 4 (a + b) x_{P} - 4 x_{P}^{2} + 4 x_{P}^{4} = 8 x_{P}^{4} - 2 x_{P}^{2} - 4 (a + b) x_{P}^{3} + (a + b) x_{P}$ $\Rightarrow 4 x_{P}^{4} - 4 (a + b) x_{P}^{3} + 2 x_{P}^{2} - 3 (a + b) x_{P} + 4 a b = 0$ $\Rightarrow x_{P}^{4} - (a + b) x_{P}^{3} + \frac{1}{2} x_{P}^{2} - \frac{3}{4} (a + b) x_{P} + a b = 0$ $x_{P} = . . . .$
	Commented by ajfour last updated on 01/Jul/19

	$o k a y s i r . T h a n k s .$
	Commented by MJS last updated on 01/Jul/19

	$please check, I get a different equation and$ $for a = 3 b = 5 our equations do not have any$ $common solution$
	Commented by mr W last updated on 01/Jul/19

	$t h a n k y o u s i r! a t y p o l e d t o w r o n g$ $e q u a t i o n i n m y w o r k i n g . n o w {i t}^{'} s f i x e d .$
	Commented by MJS last updated on 01/Jul/19

	${you}^{'} re welcome$ $I thought there must be a typo because your$ $path seemed right$
	Answered by MJS last updated on 01/Jul/19

	$P = (\begin{matrix} x \\ x^{2} \end{matrix})$ $∣ A P ∣ + ∣ B P ∣= \sqrt{x^{4} + x^{2} - 2 a x + a^{2}} + \sqrt{x^{4} + x^{2} - 2 b x + b^{2}}$ $\frac{d}{d x} [\sqrt{x^{4} + x^{2} - 2 a x + a^{2}} + \sqrt{x^{4} + x^{2} - 2 b x + b^{2}}] = 0$ $\frac{2 x^{3} + x - a}{\sqrt{x^{4} + x^{2} - 2 a x + a^{2}}} + \frac{2 x^{3} + x - b}{\sqrt{x^{4} + x^{2} - 2 b x + b^{2}}} = 0$ $after squaring and transforming we get$ $x^{4} - (a + b) x^{3} + \frac{1}{2} x^{2} - \frac{3 (a + b)}{4} x + a b = 0$
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