Angle between normal vector of ellipse and the major-axis.

I am trying to derive the angle made between the major or x-axis and the normal vector of an ellipse of general shape $x = a\cos(t),y=b\sin(t)$ with the parameter $t$ reffering to Ellipse in polar coordinates. I need to solve it for any angle $t$. From standard reasoning I find the normal vector by its definition and checked it with the page on mathworld from wolfram and works well. Then since I know 2 points, namely a point ON the shape and a point on the normal vector I derive the angle of interest to be $ \tan(\phi)=\frac{a}{b}\tan(t)$ Derivation. However this is very similar to the polar angle namely its simply the term a and b flipped. But when thinking about it I keep getting confused, am I correct or do I need the polar angle? If so where did I go wrong?

I also found Normal to Ellipse and Angle at Major Axis but this page confused me a bit, one idea I had was they use the polar angle vs the angle I am in need of ($\phi$) then I would indeed get by combing $t = \tan^{-1}(\frac a b \tan(\theta))$ and $\phi=\tan^{-1}(\frac{a}{b}\tan(t))$ $$ \phi=\tan^{-1}\bigg(\frac{a}{b}\tan(\tan^{-1}\big(\frac a b \tan\theta\big))\bigg) = \tan^{-1}\bigg(\frac{a^2}{b^2} \tan\theta\bigg) $$ My excuse for my rambling, I find these angles confusing...

*edit: fixed brackets

The tangent vector to the ellipse at parameter value $t$ is $(-a\sin t,b\cos t)$, so the tangent of the angle $\psi$ it makes with the positive $x$-axis is given by $$\tan\psi = -\frac ba\cot t.$$ The angle $\phi$ to the outward normal vector is $\psi-\pi/2$, so $$\tan\phi = \tan(\psi-\frac{\pi}2) = -\cot(\psi) = \frac ab\tan t,$$ as desired.