Verify that an ellipse has four vertices.

Question

Verify that an ellipse has four vertices.

The ellipse is given by $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

And I took $$x=a\cos t$$ and $$y=b \sin t$$ for $t\in [0,2\pi]$

Please can someone help me how to verify this?

Answer 1

Hint:

Consider the values $t=\frac {k\pi}2$ for $k\in \mathbb Z$ and the functions $\cos t,\sin t$. Are there any values of $k$ such that $\cos \frac {k\pi}2=0$? What about $\sin \frac {k\pi}2=0$?

Further hint:

We can separate the two partial derivatives $\partial x\over \partial t$ and $\partial y\over \partial t$ and get values satisfying either one of them according to the definition you are claiming, so consider the following:

$$\cos \frac \pi 2=\cos \frac {(2k+1)\pi}2=0$$

This covers two of your four "ellipse vertices". Where are the other two?

Answer 2

Use the fact that $$k=\frac{x'y''-x''y'}{((x')^2+(y')^2)^{\frac{3}{2}}}.$$ You'll get $$k'(t)=-\frac{3\sin(t)cos(t)(a^2-b^2)}{(a^2\sin^2t+b^2\sin^2t)^{\frac{5}{2}}}.$$ Wich is zero iff $\sin t=0$ or $\cos t=0$.