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Question Number 144534 by imjagoll last updated on 26/Jun/21

$$\mathrm{Find}\mathrm{the}\mathrm{shortest}\mathrm{distance}\mathrm{from}$$$$\mathrm{the}\mathrm{origin}\mathrm{to}\mathrm{the}\mathrm{hyperbola}$$$${\mathrm{x}}^2+8\mathrm{xy}+{7\mathrm{y}}^2=225,\mathrm{z}=0$$

Answered by liberty last updated on 26/Jun/21

$$\mathrm{we}\mathrm{must}\mathrm{find}\mathrm{the}\mathrm{minimum}\mathrm{value}$$$$\mathrm{of}{\mathrm{x}}^2+{\mathrm{y}}^2(\mathrm{the}\mathrm{square}\mathrm{of}\mathrm{the}\mathrm{distance}$$$$\mathrm{from}\mathrm{the}\mathrm{origin}\mathrm{to}\mathrm{any}\mathrm{point}\mathrm{in}$$$$\mathrm{the}\mathrm{xy}\mathrm{plane})\mathrm{subject}\mathrm{to}\mathrm{constraint}$$$${\mathrm{x}}^2+8\mathrm{xy}+{7\mathrm{y}}^2=225$$$$\mathrm{by}\mathrm{Langrange}\mathrm{multiplier}$$$$\mathrm{\varnothing }={\mathrm{x}}^2+8\mathrm{xy}+{7\mathrm{y}}^2+\lambda ({\mathrm{x}}^2+{\mathrm{y}}^2)$$$${\mathrm{\varnothing }}_{\mathrm{x}}=2\mathrm{x}+8\mathrm{y}+2\lambda \mathrm{x}=0\mathrm{or}(\lambda +1)\mathrm{x}+4\mathrm{y}=0$$$${\varphi }_{\mathrm{y}}=8\mathrm{x}+14\mathrm{y}+2\lambda \mathrm{y}=0\mathrm{or}4\mathrm{x}+(\lambda +7)\mathrm{y}=0$$$$\left|\begin{array}{c}\lambda +14\\ 4\lambda +7\end{array}\right|=0\to \{\begin{array}{l}\lambda =-9\\ \lambda =1\end{array}$$$$\mathrm{when}\lambda =-9\mathrm{we}\mathrm{get}\mathrm{y}=2\mathrm{x}\mathrm{and}$$$$\mathrm{substitution}\mathrm{in}{\mathrm{x}}^2+8\mathrm{xy}+{7\mathrm{y}}^2=225$$$$\mathrm{yields}{\mathrm{x}}^2=5,{\mathrm{y}}^2=20\mathrm{so}\mathrm{the}\mathrm{shortest}$$$$\mathrm{distance}\mathrm{is}\sqrt{5+20}=5$$$$\mathrm{when}\lambda =1\mathrm{give}\mathrm{x}=-2\mathrm{y}\mathrm{and}$$$$\mathrm{substitution}\mathrm{in}{\mathrm{x}}^2+8\mathrm{xy}+{7\mathrm{y}}^2=225$$$$\mathrm{yields}-{5\mathrm{y}}^2=225\mathrm{for}\mathrm{which}\mathrm{no}$$$$\mathrm{real}\mathrm{solution}\mathrm{exists}.$$

Answered by mr W last updated on 26/Jun/21

$$r=distancefrom(0,0)to(x,y)$$$$x=r\cos \theta$$$$y=r\sin \theta$$$$r^2{\cos }^2\theta +8r^2\cos \theta \sin \theta +7r^2{\sin }^2\theta =225$$$$r^2(4+4\sin 2\theta -3\cos 2\theta )=225$$$$r^2=\frac{225}{4+5\sin (2\theta -{\tan }^{-1}\frac{3}{4})}⩾\frac{225}{4+5}$$$$\Rightarrow r_{min}=\sqrt{\frac{225}{9}}=5$$

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