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

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

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

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

$$fromgraphitisseenthat(2,-4)lieson{y}^2=8x$$$$(2,-4)isnearestpointtothecircle$$$$distancebetweencentreofcircle(0,-6)and$$$$(2,-4is$$$$\sqrt{{(2-0)}^2+{(-4+6)}^2}=\sqrt{4+4}=2\sqrt{2}$$$$sominimumdistance=2\sqrt{2}-1$$$$plscheck$$

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

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

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

$$\mathrm{I}\mathrm{have}\mathrm{made}\mathrm{this}\mathrm{random}\mathrm{problem}\mathrm{myself}$$$$\mathrm{so}\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{any}\mathrm{answer}.$$

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

$$(2t^2,4t)liezon{y}^2=8x$$$$distancebetween(2t^2,4t)andthecentre(0,-6)$$$$tobeminimum$$$$s=\sqrt{{(2t^2-0)}^2+{(4t+6)}^2}$$$$s^2=4t^4+{16}^2+48t+36$$$$s^2=4(t^4+4t^2+12t+9)$$$$2s\frac{ds}{dt}=4(4t^3+8t+12)$$$$forminormax$$$$\frac{ds}{dt}=0=4t^3+8t+12$$$$t^3+2t+3=0$$$$=t^3+t^2-t^2-t+3t+3$$$$=t^2(t+1)-t(t+1)+3(t+1)$$$$=(t+1)(t^2-t+3)$$$$s\frac{ds}{dt}=2(4t^3+8t+12)$$$$s\frac{ds}{dt}=8(t^3+2t+3)$$$$s\frac{d^{2}s}{{dt}^2}+{\left({}^{}\frac{ds}{dt}\right)}^2=8(3t^2+2)$$$$putt=-1in8(3t^2+2)is40>0$$$$+veso$$$$mindistance=\sqrt{2^2+{(-4+6)}^2}fromcentre$$$$=\sqrt{8}=2\sqrt{2}$$$$requireddistance=2\sqrt{2}-1$$$$$$$$$$$$o$$

Answered by MJS last updated on 10/Sep/18

$$\mathrm{the}\mathrm{minimum}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{circle}\mathrm{and}$$$$\mathrm{the}\mathrm{parabola}\mathrm{is}\mathrm{the}\mathrm{mininum}\mathrm{distance}\mathrm{between}$$$$\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{and}\mathrm{the}\mathrm{parabola}$$$$\mathrm{minus}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}$$$$$$$$P\in par\left(\mathrm{negative}\mathrm{branch}\right):P=\left(\begin{array}{c}x\\ -2\sqrt{2x}\end{array}\right)$$$$\mathrm{center}\mathrm{of}\mathrm{circle}:C=\left(\begin{array}{c}0\\ -6\end{array}\right)$$$$\mid PC{\mid }^2=x^2+{(-6+2\sqrt{2x})}^2=x^2+8x-24\sqrt{2x}+36$$$$\frac{d}{dx}[x^2+8x-24\sqrt{2x}+36]=0$$$$2x-\frac{12\sqrt{2x}}{x}+8=0\Rightarrow$$$$\Rightarrow x(x-2)(x^2+10x+36)=0$$$$x_1=0\mathrm{is}\mathrm{a}\mathrm{local}\mathrm{maximum}$$$$x_2=2\mathrm{is}\mathrm{a}\mathrm{local}\mathrm{minimum}$$$$$$$$P=\left(\begin{array}{c}2\\ -4\end{array}\right)$$$$\mid PC\mid =\sqrt{x^2+8x-24\sqrt{2x}+36}=2\sqrt{2}$$$$\mathrm{minimum}\mathrm{distance}=\mid PC\mid -r=2\sqrt{2}-1$$

Commented by ajfour last updated on 09/Jun/18

$$Great!Sir.$$$$(butwhydidyoudeleteyourpost$$$$ofseveralintegrationquestions?)$$

Commented by MJS last updated on 09/Jun/18

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

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