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I would like to get the topmost (bottommost depending on the curve) point of an arc which means the point where the tangent is 0.

so we know that $y = c_y + r_y\sin(\alpha)\cos(t) + r_x\cos(\alpha)\sin(t)$ where $\alpha$ is the x-axis rotation value of the ellipse

we know that the tangent gives $dy/dt = 0$ (if I am not mistaken) which mean $-r_y\sin(\alpha)\sin(t) + r_x\cos(\alpha)\cos(t) = 0$

which gives $$t = \arctan{r_x\over r_y\tan(\alpha)}$$ of course $\cos(t)$ should be different from 0

but this doesn't seem to be the right result

here is desmos the A value is alpha

Intelligenti pauca
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  • @Intelligentipauca I got the equation of a rotated ellipse from here https://math.stackexchange.com/a/434482 also I have drawn the tangent in the Desmos instead of the point –  Oct 25 '20 at 18:25
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    On looking more carefully, your equation for $y$ here should have been: $$y=c_y-r_x\cos t\sin\alpha+r_y\sin t\cos\alpha.$$ – Intelligenti pauca Oct 25 '20 at 18:46
  • @Intelligentipauca thanks I completely missed that mistake –  Oct 25 '20 at 19:24

1 Answers1

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so first we have $y = c_y + r_x\sin(\alpha)\cos(t) + r_y\cos(\alpha)\sin(t)$

which gives $$t = \arctan{r_y\over r_x\tan(\alpha)}$$

after that, we should calculate the following (unrotated ellipse equation)

$u=a\cdot\cos\left(\arctan\left(\frac{b}{a\cdot\tan\left(\alpha\right)}\right)\right)$

$v=b\cdot\sin\left(\arctan\left(\frac{b}{a\cdot\tan\left(\alpha\right)}\right)\right)$

and lastly, we rotate the point (u, v) by $\alpha$

$x=u\cdot\cos\left(\alpha\right)-v\cdot\sin\left(\alpha\right)$

$y=u\cdot\sin\left(\alpha\right)+v\cdot\cos\left(\alpha\right)$

the wanted point is (x, y)

desmos