This is a reference to this question: Converting a rotated ellipse in parametric form to cartesian form.
In the answer, it is posted that to find the extreme points of an ellipse, the distance function:
$$R_1(t) = x^2(t) + y^2(t)$$
should be maximised.
My question is that why not instead:
$$R_2(t) = \sqrt{x^2(t) + y^2(t)}$$
So does this imply that maximizing $R_1(t)$ gives the same result as $R_2(t)$? And if so why?
Observe that $R_1(t)$ and $R_2(t)$ are both always non-negative, and $R_1(t) = (R_2(t))^2.$
Also note that if $a > b > 0$, then $a^2 > b^2 > 0$. Conversely, if $a > 0$, $b > 0$, and $a^2 > b^2$, then $a > b > 0$.
So if you find $t_1$ such that $R_1(t_1) \geq R_1(t)$ for all values of $t$, then $R_2(t_1) \geq R_2(t)$ for all values of $t$. Likewise, if you find $t_2$ such that $R_2(t_2) \geq R_2(t)$ for all $t$, then $R_1(t_2) \geq R_1(t)$ for all $t$.
So a maximum of $x^2(t) + y^2(t)$ is also a maximum of $\sqrt{x^2(t) + y^2(t)}$, and vice versa. But of the two expressions, $x^2(t) + y^2(t)$ is easier to work with.