Location of an arbitrary point of an ellipse

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Given this ellipse equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $(a>b>0)$ and $c:=\sqrt{a^2-b^2}$ aswell as the focal points $F=(c,0)$ and $F'=(-c,0)$,

why can we say without loss of generality, that an arbitrary point $P(x_0,y_0)$

lies in the first quadrant of the coordinate system and $x_0,y_0>0$?

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It would have been more helpful if you presented the question which contains in its solution this supposition.

Anyway, it is because the other cases can be done in a similar manner due to symmetry.

$1)$ The part of the ellipse in the $2$nd quadrant is symmetric to that in the first quadrant with respect to the $y$-axis.

$2)$ The part in the third quadrant is the symmetric of that in the first w.r.t the origin.

$3)$ The part in the fourth quadrant is symmetric to that in the first with respect to the $x$-axis.