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Question Number 83520 by john santu last updated on 03/Mar/20

$$\mathrm{closest}\mathrm{distance}\mathrm{point}(6,9)$$$$\mathrm{to}\mathrm{curve}\mathrm{y}={\mathrm{x}}^2-12\mathrm{x}+32.$$$$\mathrm{mister}\mathrm{w}\mathrm{your}\mathrm{method}\mathrm{can}\mathrm{be}$$$$\mathrm{applied}\mathrm{to}\mathrm{this}\mathrm{problem}?$$

Answered by john santu last updated on 03/Mar/20

$$\Rightarrow \mathrm{y}={(\mathrm{x}-6)}^2-4$$$${(\mathrm{x}-6)}^2=\mathrm{y}+4$$$$\mathrm{the}\mathrm{distance}\mathrm{d}=\sqrt{{(\mathrm{x}-6)}^2+{(\mathrm{y}-9)}^2}$$$${\mathrm{d}}^2=\mathrm{y}+4+{\mathrm{y}}^2-18\mathrm{y}+81$$$${\mathrm{y}}^2-17\mathrm{y}+85-{\mathrm{d}}^2=0$$$$\mathrm{tangency}\mathrm{\Delta }=0$$$${17}^2-4(85-{\mathrm{d}}^2)=0$$$$\mathrm{d}=\sqrt{\frac{340-289}{4}}=\frac{\sqrt{51}}{2}$$

Commented by jagoll last updated on 03/Mar/20

$$\mathrm{compare}\mathrm{with}\mathrm{differential}$$$$\mathrm{d}=\sqrt{{(\mathrm{x}-6)}^2+{(\mathrm{y}-9)}^2}$$$$=\sqrt{\mathrm{y}+4+{(\mathrm{y}-9)}^2}$$$$\mathrm{let}\mathrm{f}\left(\mathrm{y}\right)=\mathrm{y}+4+{(\mathrm{y}-9)}^2$$$${\mathrm{f}}^{\prime }\left(\mathrm{y}\right)=1+2(\mathrm{y}-9)=0$$$$\mathrm{y}-9=-\frac{1}{2}\Rightarrow \mathrm{y}+4=13-\frac{1}{2}$$$$\mathrm{y}+4=\frac{25}{2}\Rightarrow {\mathrm{d}}_{\min }=\sqrt{\frac{25}{2}+\frac{1}{4}}$$$${\mathrm{d}}_{\min }=\frac{\sqrt{51}}{2}$$

Commented by mr W last updated on 03/Mar/20

$$good!$$

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