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?
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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 $$A=\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}={\rm const}$$ $$x^2+2y^2=1\Rightarrow y^2=\frac{1-x^2}{2}$$ $$(x-c)^2+y^2=\frac{x^2-2x\sqrt{2}+2}{2}=\frac{(x-\sqrt{2})^2}{2}$$ $$(x+c)^2+y^2=\frac{x^2+2x\sqrt{2}+2}{2}=\frac{(x+\sqrt{2})^2}{2}$$ $$A=\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=\frac{|x-\sqrt{2}|+|x+\sqrt{2}|}{\sqrt{2}}$$ $$x^2=1-2y^2\leq 1\Rightarrow -1\leq x\leq 1\Rightarrow A=\frac{\sqrt{2}-x+x+\sqrt{2}}{\sqrt{2}}=2$$
If you want a less rigorous way, you can place two thumbtacks at the foci of an ellipse, and then wrap a string around a pencil and the thumbtacks. Pulling the string taught and tracing gives you the ellipse.
On the other hand, you can think of a circle as a special form of an ellipse where the foci are equal. In which case, it is obvious why the "sum" to the "two foci" is constant.