I am searching for a tangent (or just it's angle) to an ellipse at a specific point on the ellipse (or it's angle to the center of the ellipse).
The equation of the ellipse is $\frac{x^2}{\text{a}^2} + \frac{y^2}{\text{b}^2} = 1$.
a, b, $a$ and the point of the intersection are given and I search for
$b$, the angle of the tangent.
I tried a geometic approach which results a almost accurate angle.
I just take a line at two points, one degree left and right to 𝑎 and recieve the angle 𝑏 by using the arc tan of the points.
Is there a more accurate way?
Best Answer
The slope of the tangent to the ellipse at a point is just the derivative. Using implicit differentiation we get that $$ \frac{2x}{a^2} + \frac{2y}{b^2} \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{xb^2}{ya^2} $$ And recalling the point-slope formula for a line, if $(x_0, y_0)$ is some point on the ellipse, then the tangent line at that point is $$ y = -\frac{x_0b^2}{y_0a^2}(x-x_0) + y_0 $$