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Question Number 83445 by naka3546 last updated on 02/Mar/20

$$Thenearestdistanceof(0,3)tocurve:y=6-x^{2}is...$$

Commented by MJS last updated on 02/Mar/20

$$P=\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}x\\ 6-x^2\end{array}\right);{\mathrm{distance}}^2\mathrm{is}{x}^2+{(3-x^2)}^2=x^4-5x^2+9;\frac{d}{dx}[x^4-5x^2+9]=0\iff x(x^2-\frac{5}{2})=0\Rightarrow x=0\vee x=\pm \frac{\sqrt{10}}{2}\Rightarrow \min \mathrm{at}x=\pm \frac{\sqrt{10}}{2}\Rightarrow \min \mathrm{dist}.=\frac{\sqrt{11}}{2}$$

Answered by john santu last updated on 02/Mar/20

$$\mathrm{d}=\sqrt{{\mathrm{x}}^2+{(3-{\mathrm{x}}^2)}^2}$$$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+{(3-{\mathrm{x}}^2)}^2$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=2\mathrm{x}-2.2\mathrm{x}(3-{\mathrm{x}}^2)=0$$$$2\mathrm{x}(1-2(3-{\mathrm{x}}^2))=0$$$$\Rightarrow \mathrm{x}=0\wedge {\mathrm{x}}^2=\frac{5}{2}$$$$\mathrm{for}\mathrm{x}=0\Rightarrow \mathrm{d}=\sqrt{9}=3=\sqrt{\frac{12}{4}}$$$$\mathrm{for}{\mathrm{x}}^2=\frac{5}{2}\Rightarrow \mathrm{d}=\sqrt{\frac{5}{2}+\frac{1}{4}}=\sqrt{\frac{11}{4}}$$$${\mathrm{d}}_{\min }=\frac{1}{2}\sqrt{11}\Leftarrow \mathrm{the}\mathrm{answer}$$

Answered by mr W last updated on 02/Mar/20

$$y=6-x^2$$$$x^2+{(y-3)}^2=d^2$$$$y^2-7y+15-d^2=0$$$$tangency:$$$$\mathrm{\Delta }=7^2-4(15-d^2)=0$$$$\Rightarrow d=\sqrt{15-\frac{49}{4}}=\frac{\sqrt{11}}{2}$$

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

$$\mathrm{what}\mathrm{is}\mathrm{tangency}\mathrm{sir}?$$

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

$$ifthecircle{x}^2+{(y-3)}^2=d^{2}tangents$$$$theparabolay=6-x^2,thenradiusd$$$$ofthecircleisthesmallestdistance$$$$fromthepoint(0,3)tocurvey=6-x^2.$$$$whenbothcurvestangenteachother,$$$$theequation{y}^2-7y+15-d^2=0has$$$$oneandonlyonesolution,thisis$$$$what‘‘{tangency}^{″}means.$$

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