<s>[INST] Connection between thinking of an ellipse as a squished circle and the formal definition If I think of an ellipse fundamentaly as a squished circle. For exemple, I have a initial circle $x^2+y^2=1$ and I morph into $x^2+ 2y^2=1$.


How can I see, intuitively, that for all points on the curve, the sum of the two distances to the focal points is a constant like the formal definition of an ellipse demands?
 [/INST] What do we have?
$$x^2+2y^2=1$$
$$\frac{x^2}{1^2}+\frac{y^2}{(1/\sqrt{2})^2}=1$$
$$a=1, b=\frac{1}{\sqrt{2}}$$
$$c=\sqrt{a^2-b^2}=\frac{1}{\sqrt{2}}$$
We need to show
$$A=\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}={\rm const}$$
$$x^2+2y^2=1\Rightarrow y^2=\frac{1-x^2}{2}$$
$$(x-c)^2+y^2=\frac{x^2-2x\sqrt{2}+2}{2}=\frac{(x-\sqrt{2})^2}{2}$$
$$(x+c)^2+y^2=\frac{x^2+2x\sqrt{2}+2}{2}=\frac{(x+\sqrt{2})^2}{2}$$
$$A=\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=\frac{|x-\sqrt{2}|+|x+\sqrt{2}|}{\sqrt{2}}$$
$$x^2=1-2y^2\leq 1\Rightarrow -1\leq x\leq 1\Rightarrow A=\frac{\sqrt{2}-x+x+\sqrt{2}}{\sqrt{2}}=2$$
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