# Length of a chord parallel to the minor axis at a distance $d$ on a rotated ellipse

In this old question an equation was posted for something similar: Equation for the length of a chord parallel to either the minor or major axis in an ellipse

Anybody knows from where this equation came and if it would work with rotated ellipses too?

Thanks!

Consider the equation for an axis-aligned ellipse centered at origin: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ Here, $a$ is the semi-major axis aligned to the $x$ axis, and $b$ is the semi-major axis aligned to the $y$ axis.
Solving for $x$ you get $$x = \pm \sqrt{ a^2 - \frac{ y^2 a^2 }{ b^2 } } = \pm a \sqrt{ 1 - \left(\frac{y}{b}\right)^2 }$$ and solving for $y$ you get $$y = \pm \sqrt{ b^2 - \frac{ x^2 b^2 }{ a^2 } } = \pm b \sqrt{ 1 - \left(\frac{x}{a}\right)^2 }$$
The "width" of the ellipse (length of a horizontal chord) at $y$ is $\lvert x-(-x)\rvert$, i.e. $$L_x(y) = 2 a \sqrt{ 1 - \left(\frac{y}{b}\right)^2 }$$ and the "height" of the ellipse (length of a vertical chord) at $x$ is $\lvert y-(-y)\rvert$, i.e. $$L_y(x) = 2 b \sqrt{ 1 - \left(\frac{x}{a}\right)^2 }$$
Rotating and translating (moving) the ellipse does not change its shape or the semi-major axes, so this does apply to rotated and translated ellipses too. That is, the equation of a general ellipse is just the first equation rotated and translated. Conversely, any general ellipse can be translated to origin and rotated to have its axes parallel to $x$ and $y$ axes. This is why the above applies to all ellipses.
Let $A$ be the major axis, and $B$ the minor axis, so that $a=A/2$ is the semi-major axis, and $b=B/2$ is the semi-minor axis. (Colloquially, $A$ and $B$ are "diameters", whereas $a$ and $b$ are radii.)
At distance $d$ from the minor axis, the length of the chord parallel to the minor axis is $$L_A(d) = 2b\sqrt{1-\left(\frac{d}{a}\right)^2} = B \sqrt{1 - \left(\frac{2d}{A}\right)^2}$$ Similarly, the length of the chord parallel to the major axis at distance $d$ is $$L_B(d) = 2a\sqrt{1-\left(\frac{d}{b}\right)^2} = A \sqrt{1 - \left(\frac{2d}{B}\right)^2}$$