Relation of ellipse semi-axes with rotation angle and projection length In the following setup, assume $w$ (length of the projection of the ellipse) and $\theta$ (the rotation angle) are known. I want to know what equation(s) do I have here that helps me to derive the semi-axes ($a$ and $b$) of the ellipse based on $w$ and $\theta$.
I know the semi-axes are not unique in this problem but only the relation is needed.


$$\large 4(a^2\sin^2\theta+b^2\cos^2\theta)=w^2$$
Equation of an ellipse is:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Diffrentiate:
$$\frac{dy}{dx}=-\frac{b^2}{a^2}.\frac{x}{y}$$
Polar form of ellipse:
$$P(\phi)\equiv(a\cos\phi,b\sin\phi)$$
Slope of tangent in polar form:
$$m=-\frac ba\cot\phi$$
Equation of tangent:
$$\frac xa\cos\theta+\frac yb \sin\theta=1$$
Distance between two parallel lines in form:
$$ax+by+c=0\\ax+by+c'=0$$
is:
$$d=\frac{|c-c'|}{\sqrt{a^2+b^2}}$$


*

*Firstly those two lines are tangent at two points, and they are parallel too. So in the parametric form the slope of tangent at these two points – let them be called $P(\theta_1),Q(\theta_2)$ – must be equal.


*

*slope $\propto\cot\phi$, where $\phi$ is parametric angle.
$$\implies \cot \theta_1=\cot\theta_2\implies \theta_1=\pi+\theta_2\quad\text{since}\quad\theta_1\ne\theta_2$$


*Secondly the distance between the two tangents / two parallel lines must be $w$:


*

*Tangent Lines are $$\frac xa\cos\theta_1+\frac yb\sin\theta-1=0\\\frac xa\cos\theta+\frac yb\sin\theta+1=0$$ (after putting $\theta_2=\pi+\theta_1$). So taking distance between them:
$$\frac{|1-(-1)|}{\sqrt{\frac{\cos^2\theta_1}{a^2}+\frac{\sin^2\theta_1}{b^2}}}=w$$


*Thirdly slope of tangents at these points is $\theta$ so:
$$-\frac{b^2}{a^2}\cot\theta_1=\tan\theta$$
Try the rest yourself.
