I have an oval/ellipse, $x^2/a^2 + y^2/b^2 = 1$. The ellipse is always centered at $(0,0)$.
I have a point on that ellipse, $(x, y)$.
I'm looking for the angle of the tangent at that point relative to the x axis (or y axis, I can work from that).
So I know I need to find $y=mx+b_1$.
I'm 95% certain that the $b$ from the ellipse is not the same as the $b_1$ from the line, the use of $b$ is just coincidental. I think $arctan(m)$ is the actual angle, but I haven't figured $m$ yet.
As far as I can figure,
$m = 0$ at $y = \pm b$ and $x = 0$ because it's a flat line, so then:
$m = \infty$ at $x = \pm a$ and $y = 0$ and I have no clue what to do with that.
Everything I'm finding seems really overcomplicated and a bit confusing, I vaguely recall this being a straightforward derivative.
This is being applied in a programming situation by someone who barely remembers algebra (me), so simple is best. Hold my hand like I haven't really done this stuff in decades.
If you differentiate the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ with respect to $x$, using implicit differentiation, you get $$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$$ which simplifies to $$\frac{dy}{dx}=-\frac{b^2}{a^2}\frac{x}{y}$$ So the slope of the tangent is just $$m=-\frac{b^2}{a^2}\frac{x}{y}$$ and the angle it makes with the $x$-axis is, as you say, $\arctan(m)$.