A need to show, in $\mathbb{R}^{2}$, a metric where a Ellipse is a Sphere and I got no idea on how to proceed. Thanks!
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2try: $ d((x_1,y_1), (x_2,y_2)) = \sqrt{\frac{(x_1-x_2)^2}{a^2} + \frac{(y_1-y_2)^2}{b^2}} $ ? – Loutcho Glotuk Oct 08 '20 at 18:29
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Thats nice, added up with the hint of N. S. it may solve my problem. I've made a comment in his answer. Thanks! – big_GolfUniformIndia Oct 08 '20 at 20:25
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Hint A sphere in your metric is $$d((x,y),(x_0,y_0))=R$$
You need this to become the equation of an ellipse. What is the equation of an ellipse?
N. S.
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Ok, if I make just like Loutcho did in the comments session: define $d((x,y),(x_{1},y_{1})) = \sqrt{\frac{(x-x_{1})^{2}}{a^2} + \frac{(y-y_{1})^{2}}{b^2}}$, then $\frac{(x-x_{1})^{2}}{a^2} + \frac{(y-y_{1})^{2}}{b^2} = R^{2}$, therefore, it is an ellipse equation. Then, any point in the ellipse will lie in the Sphere of radius $R$. Is it correct? Thanks! – big_GolfUniformIndia Oct 08 '20 at 20:23
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2@GuilhermeNetto Yes. The point I was trying to make is that you could have easily figure the answer by yourself: You need $d((x,y),(x_1,y_1))=R$ to become $$\frac{(x-x_{1})^{2}}{a^2} + \frac{(y-y_{1})^{2}}{b^2} = R^{2}$$ THis equation is $$\sqrt{\frac{(x-x_{1})^{2}}{a^2} + \frac{(y-y_{1})^{2}}{b^2} }= R$$ and you need $$d((x,y),(x_1,y_1))=R$$ So what should $d((x,y),(x_1,y_1))$ be? – N. S. Oct 08 '20 at 21:51