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Question Number 97115 by bobhans last updated on 06/Jun/20

$$\mathrm{find}\mathrm{the}\mathrm{points}\mathrm{on}\mathrm{hyperbola}{\mathrm{x}}^2-{\mathrm{y}}^2=2$$$$\mathrm{closest}\mathrm{to}\mathrm{point}(0,1)$$

Answered by mr W last updated on 06/Jun/20

$$x^2+{(y-1)}^2=d^2$$$$y^2+2+{(y-1)}^2=d^2$$$$2y^2-2y+3-d^2=0$$$$\mathrm{\Delta }={(-2)}^2-4\times 2\times (3-d^2)=0$$$$2d^2=5$$$$\Rightarrow d=\frac{\sqrt{10}}{2}$$$$y=\frac{2}{2\times 2}=\frac{1}{2}$$$$x=\pm \sqrt{{\left(\frac{1}{2}\right)}^2+2}=\pm \frac{3}{2}$$$$pointsoncurve:$$$$(\pm \frac{3}{2},\frac{1}{2})$$

Commented by bobhans last updated on 06/Jun/20

$$\mathrm{at}\mathrm{point}(\pm \frac{3}{2},\frac{1}{2})?$$

Commented by bobhans last updated on 06/Jun/20

$$\mathrm{oo}\mathrm{yes}.$$

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