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I am currently reading the topic alluded to in the title of this thread. In my textbook, after the equation has been derived, $\Large\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, it says by finding the x-intercepts, $(a,0),(-a,0)$, you've also find the vertices. How do I know that these intercepts are also the vertices?

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2 Answers 2

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Since

$$y=\pm \frac{b}{a}\sqrt{a^2-x^2}$$

it is clear that it must be $\,a^2-x^2\geq 0\Longleftrightarrow |x|\leq a\,$ for the function to be defined, and thus the vertices are in the above interval's extreme points, $\,(\pm a\,,\,0)\,$

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Any affine conic can be put into such canonical equations by a suitable transformation of coordinates, yours is the case of an ellipse with semi-axes of length $a$ and $b$ aligned with the axes $x$ and $y$. To see that the interception with $x=0$ and $y=0$ gives you the vertices $(a,0),(-a,0)$ and $(0,b),(0,-b)$, you may solve one of the variables, for example $$y=\pm |b|\sqrt{1-\frac{x^2}{a^2}},$$ and see that those functions have absolute maximum and minimum at $(0,|b|),(0,-|b|)$ respectively; indeed, the square root is only defined for $x^2/a^2\leq1$ and attains its maximum value at $x=0$. Similarly solving for $x(y)$ you get the vertices of the other axis at $(|a|,0),(-|a|,0)$.

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