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**Soroban** Hello, MegaVortex7!

We have: .$\displaystyle y \;=\;\sqrt{2x-2x^2}$

Square: .$\displaystyle y^2 \;=\;2x - 2x^2 \quad\Rightarrow \quad 2x^2 - 2x + y^2 \:=\:0$

Complete the square: .$\displaystyle 2(x^2 - x) + y^2 \:=\:0 \quad\Rightarrow\quad 2\left(x^2 - x + \tfrac{1}{4}\right) + y^2 \:=\:\tfrac{1}{2}$

. . $\displaystyle 2\left(x - \tfrac{1}{2}\right)^2 + y^2 \:=\:\tfrac{1}{2} \quad\Rightarrow\quad \frac{(x-\frac{1}{2})^2}{\frac{1}{4}} + \frac{y^2}{\frac{1}{2}} \;=\;1$

This is an ellipse; its major axis is vertical.

. . The semimajor axis is: .$\displaystyle a \,=\,\tfrac{\sqrt{2}}{2}$

. . The semiminor axis is: .$\displaystyle b \,=\,\tfrac{1}{2}$

And the graph is the *upper half* of this ellipse.