Algebra Craze 3

Let $|m| > \sqrt{2}$ .

$\begin{aligned} x+y &=& mX - 1 \\ xy &=& \dfrac{1}{2}((m^2 - 1)X^2 - mX - 2) \end{aligned}$

$\implies$

$\begin{aligned} (x+y)^2 &=& m^2X^2 - 2mX + 1\\ 2xy &=& (m^2 - 1)X^2 - mX - 2 \end{aligned}$

$\implies (x + y)^2 - 2xy = x^2 + y^2 = X^2 - mX + 3$ .

$x = \dfrac{(m^2 - 1)X^2 - mX - 2}{2y} \implies 2y^2 - 2(mX - 1)y + ((m^2 - 1)^2 - mX - 2) = 0$ which has real value solutions whenever the discriminant $D = 20 - (4m^2 - 8)X^2 \geq 0 \implies |X|^2 \leq \dfrac{5}{m^2 - 2} \implies |X| \leq \sqrt{\dfrac{5}{m^2 - 2}} \implies$

$x^2 + y^2 = \dfrac{5}{m^2 - 2} - \sqrt{\dfrac{5}{m^2 - 2}}m + 3 = 3 - \sqrt{\dfrac{5}{m^2 - 1}}m + 1 \implies \dfrac{5}{m^2 - 2} = 1 \implies m = \pm\sqrt{7}$

For $m > \sqrt{2} \implies m = \sqrt{7} \implies x^2 + y^2 = (4 - \sqrt{7})$ and $|X| \leq 1$ .

Using $m = \sqrt{7}$ and $X = 1 \implies$

$\begin{aligned} x+y &=& \sqrt{7} - 1 \\ y &=& \dfrac{4 - \sqrt{7}}{2x} \end{aligned}$

$\implies 2x^2 - 2(\sqrt{7} - 1)x + 4 - \sqrt{7} = 0 \implies \boxed{x_{0} = \dfrac{\sqrt{7} - 1}{2} \implies y_{0} = \dfrac{4 - \sqrt{7}}{\sqrt{7} - 1} = \dfrac{\sqrt{7} - 1}{2}}$

and

$y = \dfrac{4 - \sqrt{7}}{2x} \implies \dfrac{dy}{dx}|_{x = \dfrac{\sqrt{7} - 1}{2}} =$ $-\dfrac{4 - \sqrt{7}}{2x^2}|_{x = \dfrac{\sqrt{7} - 1}{2}} = \dfrac{-2(4 - \sqrt{7})}{(\sqrt{7} - 1)^2} =$

$-\dfrac{(4 - \sqrt{7})(\sqrt{7} + 1)}{3(\sqrt{7} - 1)} = \dfrac{-3(\sqrt{7} - 1)}{3(\sqrt{7} - 1)} = -1$

$\implies y - (\dfrac{\sqrt{7} -1}{2}) = -1(x - (\dfrac{\sqrt{7} - 1}{2}) \implies \boxed{x + y = \sqrt{7} - 1}$ .

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For $m < -\sqrt{2} \implies m = -\sqrt{7}$ and $X = 1 \implies$

$\begin{aligned} x+y &=& -(\sqrt{7} + 1) \\ y &=& \dfrac{4 + \sqrt{7}}{2x} \end{aligned}$

$\implies 2x^2 + 2(\sqrt{7} + 1)x + 4 + \sqrt{7} = 0 \implies \boxed{x^{*}_{0} = -\dfrac{\sqrt{7} + 1}{2} \implies y^{*}_{0} = -\dfrac{\sqrt{7} + 1}{2}}$

and

$y = \dfrac{4 + \sqrt{7}}{2x} \implies \dfrac{dy}{dx}|_{x = -\dfrac{\sqrt{7} + 1}{2}} =$ $-\dfrac{4 + \sqrt{7}}{2x^2}|_{x = -\dfrac{\sqrt{7} + 1}{2}}$ $= \dfrac{-2(4 + \sqrt{7})}{(\sqrt{7} + 1)^2} =$

$-\dfrac{(4 + \sqrt{7})(\sqrt{7} - 1)}{3(\sqrt{7} + 1)} = \dfrac{-3(\sqrt{7} + 1)}{3(\sqrt{7} +1)} = -1$

$\implies \boxed{x + y = -(\sqrt{7} + 1)}$ .

Using points $A (x_{0},y_{0})$ and $B (x^{*}_{0},y^{*}_{0}) \implies$ the distance $d = \sqrt{14} \approx \boxed{3.74165739}$ .