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

The nearest distance of (0, 3) to curve : y = 6 − x^2 is ...

$${The}\:\:{nearest}\:\:{distance}\:\:{of}\:\:\left(\mathrm{0},\:\mathrm{3}\right)\:\:{to}\:\:{curve}\:\::\:\:{y}\:=\:\:\mathrm{6}\:−\:{x}^{\mathrm{2}} \:\:{is}\:\:... \\ $$

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)

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

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

d = (√(x^2 +(3−x^2 )^2 )) let f(x) = x^2 +(3−x^2 )^2 f ′(x) = 2x−2.2x(3−x^2 )=0 2x(1−2(3−x^2 )) =0 ⇒ x = 0 ∧ x^2 = (5/2) for x = 0 ⇒d = (√9) = 3 =(√((12)/4)) for x^2 = (5/2)⇒d =(√((5/2)+(1/4)))=(√((11)/4)) d_(min ) = (1/2)(√(11)) ⇐ the answer

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

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

y=6−x^2 x^2 +(y−3)^2 =d^2 y^2 −7y+15−d^2 =0 tangency: Δ=7^2 −4(15−d^2 )=0 ⇒d=(√(15−((49)/4)))=((√(11))/2)

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

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

what is tangency sir?

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

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

if the circle x^2 +(y−3)^2 =d^2 tangents the parabola y=6−x^2 , then radius d of the circle is the smallest distance from the point (0,3) to curve y=6−x^2 . when both curves tangent each other, the equation y^2 −7y+15−d^2 =0 has one and only one solution, this is what “tangency”means.

$${if}\:{the}\:{circle}\:{x}^{\mathrm{2}} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} ={d}^{\mathrm{2}} \:{tangents} \\ $$$${the}\:{parabola}\:{y}=\mathrm{6}−{x}^{\mathrm{2}} ,\:{then}\:{radius}\:{d} \\ $$$${of}\:{the}\:{circle}\:{is}\:{the}\:{smallest}\:{distance} \\ $$$${from}\:{the}\:{point}\:\left(\mathrm{0},\mathrm{3}\right)\:{to}\:{curve}\:{y}=\mathrm{6}−{x}^{\mathrm{2}} . \\ $$$${when}\:{both}\:{curves}\:{tangent}\:{each}\:{other}, \\ $$$${the}\:{equation}\:{y}^{\mathrm{2}} −\mathrm{7}{y}+\mathrm{15}−{d}^{\mathrm{2}} =\mathrm{0}\:{has} \\ $$$${one}\:{and}\:{only}\:{one}\:{solution},\:{this}\:{is} \\ $$$${what}\:``{tangency}''{means}. \\ $$

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