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Question Number 121264 by benjo_mathlover last updated on 06/Nov/20

$$\mathrm{Find}\mathrm{the}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{curve}$$$$\mathrm{x}={2\mathrm{y}}^2\mathrm{closest}\mathrm{to}\mathrm{the}\mathrm{point}$$$$(10,0)$$

Answered by liberty last updated on 06/Nov/20

$$\mathrm{Let}\mathrm{L}\mathrm{be}\mathrm{the}\mathrm{square}\mathrm{of}\mathrm{distance}\mathrm{between}$$$$(\mathrm{x},\mathrm{y})\mathrm{and}(10,0).$$$$\mathrm{L}={(\mathrm{x}-10)}^2+{\mathrm{y}}^2={({2\mathrm{y}}^2-10)}^2+{\mathrm{y}}^2$$$$\mathrm{L}={4\mathrm{y}}^4-{39\mathrm{y}}^2+100$$$$\frac{\mathrm{dL}}{\mathrm{dy}}={16\mathrm{y}}^3-78\mathrm{y}=0$$$$2\mathrm{y}({8\mathrm{y}}^2-39)=0,\mathrm{y}=0;\pm \frac{\sqrt{39}}{2\sqrt{2}}$$$$\frac{\mathrm{dL}}{\mathrm{dy}}<0\mathrm{on}(-\mathrm{\infty },-\frac{\sqrt{39}}{2\sqrt{2}})\mathrm{and}(0,\frac{\sqrt{39}}{2\sqrt{2}})$$$$\frac{\mathrm{dL}}{\mathrm{dy}}>0\mathrm{on}(-\frac{\sqrt{39}}{2\sqrt{2}},0)\mathrm{and}(\frac{\sqrt{39}}{2\sqrt{2}},\mathrm{\infty })$$$$\mathrm{When}\mathrm{y}=-\frac{\sqrt{39}}{2\sqrt{2}},\mathrm{x}=\frac{39}{4}$$$$\mathrm{and}\mathrm{y}=\frac{\sqrt{39}}{2\sqrt{2}},\mathrm{x}=\frac{39}{4}.\because \mathrm{The}\mathrm{points}\mathrm{are}$$$$(\frac{39}{4},-\frac{\sqrt{39}}{2\sqrt{2}})\mathrm{and}(\frac{39}{4},\frac{\sqrt{39}}{2\sqrt{2}})$$

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

$$saypointP(2p^2,p)$$$$1=4y\frac{dy}{dx}$$$$\frac{dy}{dx}=\frac{1}{4p}=-\frac{10-2p^2}{0-p}$$$$\Rightarrow -8p^2+40=1$$$$\Rightarrow p=\pm \frac{\sqrt{39}}{2\sqrt{2}}$$$$\Rightarrow P(\frac{39}{4},\pm \frac{\sqrt{39}}{2\sqrt{2}})$$

Commented by liberty last updated on 06/Nov/20

$$\mathrm{it}\mathrm{should}\mathrm{be}1=4\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}$$$$$$

Commented by mr W last updated on 06/Nov/20

$$yes.thanks!$$

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