Find the point on an ellipse by angle. How do I find the point on the ellipses at 45'. I found this, which answers part of it, but I need to know how to calculate for (x,y) at 45'. I could also use a good explanation for the t variable.  Thanks...
x = a cos t 
y = b sin t

 A: Suppose you have  a $\theta$ value which represents the angle you want to find the $(x,y)$ value of the ellipse at. We can represent a line passing through the origin given $\theta$ in the form of $y=mx$, where $m$ is the slope of the line. The slope of the line will simply be $\frac{dy}{dx}=\frac{\sin\theta}{\cos\theta}=\tan\theta$, and so the line becomes $y=(\tan\theta$)x. 
Note that your parametric equations are $x=a\cos t$,$y=b \sin t$. If you solve for $t$ and substitute, you can get $$(\frac{x}{a})^2+(\frac{y}{b})^2=1$$
Now we need to find the intersection of the ellipse and the line by plugging in $y$.
$$(\frac{x}{a})^2+(\frac{x\tan\theta}{b})^2=1\\x^2(1/a^2+\tan^2\theta/b^2)=1\\x=\pm\sqrt{\frac{1}{1/a^2+\tan^2\theta/b^2}}\\y=(\tan\theta)(\pm\sqrt{\frac{1}{1/a^2+\tan^2\theta/b^2}})$$
Note that the $\pm$ depends on what your angle is. If it is $45$ degrees, then you want both $x$ and $y$ to be positive.
A: On the 45 degree line, $x = y$, so $a \cos t = b \sin t$.  
Assuming $b \neq 0$, $\dfrac{a}{b} = \dfrac{\sin t}{\cos t} = \tan t$.  
So if $\theta = \arctan \dfrac{a}{b}$, $t = \theta$ and $t = \theta + \pi$ both work.
Apply values of $t$ to the parametric equations to get the actual $x$ and $y$ values.
