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Question Number 144995 by liberty last updated on 01/Jul/21

$$\mathrm{Find}\mathrm{the}\mathrm{maximum}\mathrm{distance}$$$$\mathrm{between}\mathrm{two}\mathrm{points}\mathrm{on}\mathrm{the}$$$$\mathrm{curve}\frac{{\mathrm{x}}^4}{{\mathrm{a}}^4}+\frac{{\mathrm{y}}^4}{{\mathrm{b}}^4}=1.$$

Answered by mr W last updated on 01/Jul/21

$$duetosymmetry,themaximum$$$$distancebetweentwopointsonthe$$$$curveistwotimesofthemaximum$$$$distancefromapointonthecurve$$$$totheorigin.$$$$pointP(x,y)$$$$distancefromP(x,y)to(0,0)isr.$$$$x=r\cos \theta$$$$y=r\sin \theta$$$$\frac{r^4{\cos }^4\theta }{a^4}+\frac{r^4{\sin }^4\theta }{b^4}=1$$$$\frac{{\cos }^4\theta }{a^2}+\frac{{\sin }^4\theta }{b^4}=\frac{1}{r^4}=\mathrm{\Phi }$$$$\frac{d\mathrm{\Phi }}{d\theta }=-\frac{4{\cos }^3\theta \sin \theta }{a^4}+\frac{4{\sin }^3\theta \cos \theta }{b^4}=0$$$$(-\frac{{\cos }^2\theta }{a^4}+\frac{{\sin }^2\theta }{b^4})\cos \theta \sin \theta =0$$$$\Rightarrow \sin \theta =0\Rightarrow \theta =0\Rightarrow r=a$$$$\Rightarrow \cos \theta =0\Rightarrow \theta =\frac{\pi }{2}\Rightarrow r=b$$$$\Rightarrow -\frac{{\cos }^2\theta }{a^4}+\frac{{\sin }^2\theta }{b^4}$$$$\Rightarrow \tan \theta =\frac{b^2}{a^2}$$$$\Rightarrow \frac{1}{r_{max}^4}=\frac{1}{a^4+b^4}$$$$\Rightarrow r_{max}=\sqrt[4]{a^4+b^4}>a,>b$$$$\Rightarrow d_{max}=2r_{max}=2\sqrt[4]{a^4+b^4}$$

Commented by mr W last updated on 01/Jul/21

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