Is phi the Actual Angle on an Ellipse?

[elegysix]

given an ellipse in vector form
r(phi)=a*cos(phi)i +b*sin(phi)j
where i and j are the unit vectors for x and y,
then y= b*sin(phi), and x = a*cos(phi).
tan(phi) = y / x ,
but y/x=(b/a)*tan(phi)
which implies 1 = b/a or b=a
which is false.

What is the deal?
doing the same for a circle leads to the true statement, 1=1.

 

[LCKurtz]

What is wrong is that in your parameterization:

r(phi)=a*cos(phi)i +b*sin(phi)j

the parameter phi is not equal to the polar coordinate angle theta unless a = b.

 

[elegysix]

I never mentioned anything about polar coordinates, or its angle theta.
Since r is in cartesian components,
y must be equal to b*sin($$\phi$$),
and x must be equal to a*cos($$\phi$$), right?
and since they are perpendicular, they form a right triangle with r as the hypotenuse... then the tangent of that angle, tan($$\phi$$) must equal y/x, no?

 

[ideasrule]

I never mentioned anything about polar coordinates, or its angle theta.
Since r is in cartesian components,
y must be equal to b*sin($$\phi$$),
and x must be equal to a*cos($$\phi$$), right?
and since they are perpendicular, they form a right triangle with r as the hypotenuse... then the tangent of that angle, tan($$\phi$$) must equal y/x, no?

phi is a parameter, not an actual angle. That is, if you choose a certain value for phi, you'll get a point (x,y) that's on the ellipse. However, if you actually plot this point and connect it to (0,0), the resulting line is NOT at an angle phi from the x-axis. There's no reason it should be, since phi is an arbitrarily picked value.