Question Number 83445 by naka3546 last updated on 02/Mar/20
$${The}\:\:{nearest}\:\:{distance}\:\:{of}\:\:\left(\mathrm{0},\:\mathrm{3}\right)\:\:{to}\:\:{curve}\:\::\:\:{y}\:=\:\:\mathrm{6}\:−\:{x}^{\mathrm{2}} \:\:{is}\:\:… \\ $$
Commented by MJS last updated on 02/Mar/20
![P= ((x),(y) ) = ((x),((6−x^2 )) ); distance^2 is x^2 +(3−x^2 )^2 =x^4 −5x^2 +9; (d/dx)[x^4 −5x^2 +9]=0 ⇔ x(x^2 −(5/2))=0 ⇒ x=0∨x=±((√(10))/2) ⇒ min at x=±((√(10))/2) ⇒ min dist. = ((√(11))/2)](https://www.tinkutara.com/question/Q83452.png)
$${P}=\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{{x}}\\{\mathrm{6}−{x}^{\mathrm{2}} }\end{pmatrix};\:\mathrm{distance}^{\mathrm{2}} \:\mathrm{is}\:{x}^{\mathrm{2}} +\left(\mathrm{3}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} ={x}^{\mathrm{4}} −\mathrm{5}{x}^{\mathrm{2}} +\mathrm{9};\:\frac{{d}}{{dx}}\left[{x}^{\mathrm{4}} −\mathrm{5}{x}^{\mathrm{2}} +\mathrm{9}\right]=\mathrm{0}\:\Leftrightarrow\:{x}\left({x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{2}}\right)=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{0}\vee{x}=\pm\frac{\sqrt{\mathrm{10}}}{\mathrm{2}}\:\Rightarrow\:\mathrm{min}\:\mathrm{at}\:{x}=\pm\frac{\sqrt{\mathrm{10}}}{\mathrm{2}}\:\Rightarrow\:\mathrm{min}\:\mathrm{dist}.\:=\:\frac{\sqrt{\mathrm{11}}}{\mathrm{2}} \\ $$
Answered by john santu last updated on 02/Mar/20
$$\mathrm{d}\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{x}^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\mathrm{2x}−\mathrm{2}.\mathrm{2x}\left(\mathrm{3}−\mathrm{x}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{2x}\left(\mathrm{1}−\mathrm{2}\left(\mathrm{3}−\mathrm{x}^{\mathrm{2}} \right)\right)\:=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{x}\:=\:\mathrm{0}\:\wedge\:\mathrm{x}^{\mathrm{2}} =\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{for}\:\mathrm{x}\:=\:\mathrm{0}\:\Rightarrow\mathrm{d}\:=\:\sqrt{\mathrm{9}}\:=\:\mathrm{3}\:=\sqrt{\frac{\mathrm{12}}{\mathrm{4}}} \\ $$$$\mathrm{for}\:\mathrm{x}^{\mathrm{2}} \:=\:\frac{\mathrm{5}}{\mathrm{2}}\Rightarrow\mathrm{d}\:=\sqrt{\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}}=\sqrt{\frac{\mathrm{11}}{\mathrm{4}}} \\ $$$$\mathrm{d}_{\mathrm{min}\:} =\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{11}}\:\Leftarrow\:\mathrm{the}\:\mathrm{answer} \\ $$
Answered by mr W last updated on 02/Mar/20
$${y}=\mathrm{6}−{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} ={d}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} −\mathrm{7}{y}+\mathrm{15}−{d}^{\mathrm{2}} =\mathrm{0} \\ $$$${tangency}: \\ $$$$\Delta=\mathrm{7}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{15}−{d}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{d}=\sqrt{\mathrm{15}−\frac{\mathrm{49}}{\mathrm{4}}}=\frac{\sqrt{\mathrm{11}}}{\mathrm{2}} \\ $$
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
$${if}\:{the}\:{circle}\:{x}^{\mathrm{2}} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} ={d}^{\mathrm{2}} \:{tangents} \\ $$$${the}\:{parabola}\:{y}=\mathrm{6}−{x}^{\mathrm{2}} ,\:{then}\:{radius}\:{d} \\ $$$${of}\:{the}\:{circle}\:{is}\:{the}\:{smallest}\:{distance} \\ $$$${from}\:{the}\:{point}\:\left(\mathrm{0},\mathrm{3}\right)\:{to}\:{curve}\:{y}=\mathrm{6}−{x}^{\mathrm{2}} . \\ $$$${when}\:{both}\:{curves}\:{tangent}\:{each}\:{other}, \\ $$$${the}\:{equation}\:{y}^{\mathrm{2}} −\mathrm{7}{y}+\mathrm{15}−{d}^{\mathrm{2}} =\mathrm{0}\:{has} \\ $$$${one}\:{and}\:{only}\:{one}\:{solution},\:{this}\:{is} \\ $$$${what}\:“{tangency}''{means}. \\ $$