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Ok this is a really silly question and I should know this, but I can't seem to figure something out:alt text

for the last step, how do they know that $0 \leq x \leq 4$? If we use the minimum value of theta, which is $-\pi/2$, and plug that into $x=\cos(\theta)$, then we get $0$, and same for $\pi/2$.

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up vote 3 down vote accepted

Because over $-\pi /2\leq \theta \leq \pi /2, 0\leq\cos{\theta}\leq1$. At $\theta =0, \cos{\theta}=1$

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Well, y^2 has to be positive, so the minimum that y^2/25 could be is 0. So, x^2/16 is 1 at the maximum => x≤4. The same argument applies for x = 0

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By your argument $x^2 \geq 0$, which does not imply that $x \geq 0$, only that $|x| \geq 0$, which is true for any $x$. The key is the domain on which $\theta$ is defined, as $x$ can be negative on the left half of the ellipse which is not contained in the parametrization. – Brandon Carter Nov 11 '10 at 5:07

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