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Question Number 37137 by rahul 19 last updated on 09/Jun/18

Find minimum distance between y^2 =8x and x^2 +(y+6)^2 =1.

$$\mathrm{Find}\:\mathrm{minimum}\:\mathrm{distance}\:\mathrm{between} \\ $$$$\mathrm{y}^{\mathrm{2}} =\mathrm{8}{x}\:{and}\:{x}^{\mathrm{2}} +\left({y}+\mathrm{6}\right)^{\mathrm{2}} =\mathrm{1}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18

from graph it is seen that(2,−4) lies on y^2 =8x (2,−4) is nearest point to the circle distance between centre of circle(0,−6) and (2,−4 is (√((2−0)^2 +(−4+6)^2 )) =(√(4+4)) =2(√2) so minimum distance=2(√2) −1 pls check

$${from}\:{graph}\:{it}\:{is}\:{seen}\:{that}\left(\mathrm{2},−\mathrm{4}\right)\:{lies}\:{on}\:{y}^{\mathrm{2}} =\mathrm{8}{x} \\ $$$$\left(\mathrm{2},−\mathrm{4}\right)\:{is}\:{nearest}\:{point}\:{to}\:{the}\:{circle} \\ $$$${distance}\:{between}\:{centre}\:{of}\:{circle}\left(\mathrm{0},−\mathrm{6}\right)\:{and}\: \\ $$$$\left(\mathrm{2},−\mathrm{4}\:{is}\right. \\ $$$$\sqrt{\left(\mathrm{2}−\mathrm{0}\right)^{\mathrm{2}} +\left(−\mathrm{4}+\mathrm{6}\right)^{\mathrm{2}} \:}\:\:=\sqrt{\mathrm{4}+\mathrm{4}}\:\:=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${so}\:{minimum}\:{distance}=\mathrm{2}\sqrt{\mathrm{2}}\:\:−\mathrm{1} \\ $$$${pls}\:{check} \\ $$

Commented by rahul 19 last updated on 09/Jun/18

Sir, how do you conclude (2,−4) is the nearest point to circle. Also sol. using parametric coordinates is welcome!

$$\mathrm{Sir},\:\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{conclude}\:\left(\mathrm{2},−\mathrm{4}\right)\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{nearest}\:\mathrm{point}\:\mathrm{to}\:\mathrm{circle}. \\ $$$$\mathrm{Also}\:\mathrm{sol}.\:\mathrm{using}\:\mathrm{parametric}\:\mathrm{coordinates} \\ $$$$\mathrm{is}\:\mathrm{welcome}! \\ $$

Commented by rahul 19 last updated on 09/Jun/18

I have made this random problem myself so i don′t have any answer.

$$\mathrm{I}\:\mathrm{have}\:\mathrm{made}\:\mathrm{this}\:\mathrm{random}\:\mathrm{problem}\:\mathrm{myself} \\ $$$$\mathrm{so}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{any}\:\mathrm{answer}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18

(2t^2 ,4t) liez on y^2 =8x distance between (2t^2 ,4t) and the centre(0,−6) to be minimum s=(√((2t^2 −0)^2 +(4t+6)^2 )) s^2 =4t^4 +16^2 +48t+36 s^2 =4(t^4 +4t^2 +12t+9) 2s(ds/dt)=4(4t^3 +8t+12) for min or max (ds/dt)=0=4t^3 +8t+12 t^3 +2t+3=0 =t^3 +t^2 −t^2 −t+3t+3 =t^2 (t+1) −t(t+1) +3(t+1) =(t+1)(t^2 −t+3) s(ds/dt)=2(4t^3 +8t+12) s(ds/dt)=8(t^3 +2t+3) s(d^2 s/dt^2 )+(^ (ds/dt))^2 =8(3t^2 +2) put t=−1 in8(3t^2 +2) is40>0 +ve so min distance =(√(2^2 +(−4+6)^2 )) from centre =(√8) =2(√2) required distance=2(√2) −1 o

$$\left(\mathrm{2}{t}^{\mathrm{2}} ,\mathrm{4}{t}\right)\:{liez}\:{on}\:{y}^{\mathrm{2}} =\mathrm{8}{x} \\ $$$${distance}\:{between}\:\left(\mathrm{2}{t}^{\mathrm{2}} ,\mathrm{4}{t}\right)\:{and}\:{the}\:{centre}\left(\mathrm{0},−\mathrm{6}\right) \\ $$$${to}\:{be}\:{minimum} \\ $$$${s}=\sqrt{\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{0}\right)^{\mathrm{2}} +\left(\mathrm{4}{t}+\mathrm{6}\right)^{\mathrm{2}} \:} \\ $$$${s}^{\mathrm{2}} =\mathrm{4}{t}^{\mathrm{4}} +\mathrm{16}^{\mathrm{2}} +\mathrm{48}{t}+\mathrm{36} \\ $$$${s}^{\mathrm{2}} =\mathrm{4}\left({t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} +\mathrm{12}{t}+\mathrm{9}\right) \\ $$$$\mathrm{2}{s}\frac{{ds}}{{dt}}=\mathrm{4}\left(\mathrm{4}{t}^{\mathrm{3}} +\mathrm{8}{t}+\mathrm{12}\right) \\ $$$${for}\:{min}\:{or}\:{max} \\ $$$$\frac{{ds}}{{dt}}=\mathrm{0}=\mathrm{4}{t}^{\mathrm{3}} +\mathrm{8}{t}+\mathrm{12} \\ $$$${t}^{\mathrm{3}} +\mathrm{2}{t}+\mathrm{3}=\mathrm{0} \\ $$$$={t}^{\mathrm{3}} +{t}^{\mathrm{2}} −{t}^{\mathrm{2}} −{t}+\mathrm{3}{t}+\mathrm{3} \\ $$$$={t}^{\mathrm{2}} \left({t}+\mathrm{1}\right)\:\:\:−{t}\left({t}+\mathrm{1}\right)\:\:\:+\mathrm{3}\left({t}+\mathrm{1}\right) \\ $$$$=\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −{t}+\mathrm{3}\right) \\ $$$${s}\frac{{ds}}{{dt}}=\mathrm{2}\left(\mathrm{4}{t}^{\mathrm{3}} +\mathrm{8}{t}+\mathrm{12}\right) \\ $$$${s}\frac{{ds}}{{dt}}=\mathrm{8}\left({t}^{\mathrm{3}} +\mathrm{2}{t}+\mathrm{3}\right) \\ $$$${s}\frac{{d}^{\mathrm{2}} {s}}{{dt}^{\mathrm{2}} }+\left(^{} \frac{{ds}}{{dt}}\right)^{\mathrm{2}} =\mathrm{8}\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{2}\right) \\ $$$${put}\:{t}=−\mathrm{1}\:{in}\mathrm{8}\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{2}\right)\:{is}\mathrm{40}>\mathrm{0} \\ $$$$+{ve}\:{so} \\ $$$${min}\:{distance}\:=\sqrt{\mathrm{2}^{\mathrm{2}} +\left(−\mathrm{4}+\mathrm{6}\right)^{\mathrm{2}} }\:{from}\:{centre} \\ $$$$=\sqrt{\mathrm{8}}\:\:=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${required}\:{distance}=\mathrm{2}\sqrt{\mathrm{2}}\:−\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$${o} \\ $$

Answered by MJS last updated on 10/Sep/18

the minimum distance between the circle and the parabola is the mininum distance between the center of the circle and the parabola minus the radius of the circle P∈par (negative branch): P= ((x),((−2(√(2x)))) ) center of circle: C= ((0),((−6)) ) ∣PC∣^2 =x^2 +(−6+2(√(2x)))^2 =x^2 +8x−24(√(2x))+36 (d/dx)[x^2 +8x−24(√(2x))+36]=0 2x−((12(√(2x)))/x)+8=0 ⇒ ⇒ x(x−2)(x^2 +10x+36)=0 x_1 =0 is a local maximum x_2 =2 is a local minimum P= ((2),((−4)) ) ∣PC∣=(√(x^2 +8x−24(√(2x))+36))=2(√2) minimum distance =∣PC∣−r=2(√2)−1

$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{parabola}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mininum}\:\mathrm{distance}\:\mathrm{between} \\ $$$$\mathrm{the}\:\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{and}\:\mathrm{the}\:\mathrm{parabola} \\ $$$$\mathrm{minus}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle} \\ $$$$ \\ $$$${P}\in{par}\:\left(\mathrm{negative}\:\mathrm{branch}\right):\:{P}=\begin{pmatrix}{{x}}\\{−\mathrm{2}\sqrt{\mathrm{2}{x}}}\end{pmatrix} \\ $$$$\mathrm{center}\:\mathrm{of}\:\mathrm{circle}:\:{C}=\begin{pmatrix}{\mathrm{0}}\\{−\mathrm{6}}\end{pmatrix} \\ $$$$\mid{PC}\mid^{\mathrm{2}} ={x}^{\mathrm{2}} +\left(−\mathrm{6}+\mathrm{2}\sqrt{\mathrm{2}{x}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{24}\sqrt{\mathrm{2}{x}}+\mathrm{36} \\ $$$$\frac{{d}}{{dx}}\left[{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{24}\sqrt{\mathrm{2}{x}}+\mathrm{36}\right]=\mathrm{0} \\ $$$$\mathrm{2}{x}−\frac{\mathrm{12}\sqrt{\mathrm{2}{x}}}{{x}}+\mathrm{8}=\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:{x}\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{36}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{0}\:\mathrm{is}\:\mathrm{a}\:\mathrm{local}\:\mathrm{maximum} \\ $$$${x}_{\mathrm{2}} =\mathrm{2}\:\mathrm{is}\:\mathrm{a}\:\mathrm{local}\:\mathrm{minimum} \\ $$$$ \\ $$$${P}=\begin{pmatrix}{\mathrm{2}}\\{−\mathrm{4}}\end{pmatrix} \\ $$$$\mid{PC}\mid=\sqrt{{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{24}\sqrt{\mathrm{2}{x}}+\mathrm{36}}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{minimum}\:\mathrm{distance}\:=\mid{PC}\mid−{r}=\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1} \\ $$

Commented by ajfour last updated on 09/Jun/18

Great! Sir. (but why did you delete your post of several integration questions?)

$${Great}!\:{Sir}. \\ $$$$\left({but}\:{why}\:{did}\:{you}\:{delete}\:{your}\:{post}\right. \\ $$$$\left.{of}\:{several}\:{integration}\:{questions}?\right) \\ $$

Commented by MJS last updated on 09/Jun/18

I made some mistakes, will post the corrected version soon

$$\mathrm{I}\:\mathrm{made}\:\mathrm{some}\:\mathrm{mistakes},\:\mathrm{will}\:\mathrm{post}\:\mathrm{the}\:\mathrm{corrected} \\ $$$$\mathrm{version}\:\mathrm{soon} \\ $$

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