if $\alpha$, $\beta$, $\gamma$, $\delta$ be the eccentric angles of four points of intersection of the ellipse and any circle,prove that $\alpha+\delta+\beta+\gamma$ is an even multiple of $\pi$ radians.
[Math] Ellipse and circle
conic sections
Related Solutions
The equation to the chord of the ellipse joining two points with eccentric angles $\alpha$ and $\beta$ is $\frac{x}{a}\cos \frac{\alpha+\beta}{2}+\frac{y}{b}\sin \frac{\alpha+\beta}{2}=\cos\frac{\alpha-\beta}{2}$
Similarly,The equation to the chord of the ellipse joining two points with eccentric angles $\gamma$ and $\delta$ is $\frac{x}{a}\cos \frac{\gamma+\delta}{2}+\frac{y}{b}\sin \frac{\gamma+\delta}{2}=\cos\frac{\gamma-\delta}{2}$
Since,first chord passes through $(d,0)$ and the second chord passes through $(-d,0)$.
$\frac{d}{a}\cos \frac{\alpha+\beta}{2}=\cos\frac{\alpha-\beta}{2}$ and
$\frac{-d}{a}\cos \frac{\gamma+\delta}{2}=\cos\frac{\gamma-\delta}{2}$
$\Rightarrow \frac{\cos\frac{\alpha-\beta}{2}}{\cos\frac{\alpha+\beta}{2}}=-\frac{\cos\frac{\gamma-\delta}{2}}{\cos\frac{\gamma+\delta}{2}}$
$\Rightarrow \frac{\cos\frac{\alpha}{2}\cos\frac{\beta}{2}+\sin\frac{\alpha}{2}\sin\frac{\beta}{2}}{\cos\frac{\alpha}{2}\cos\frac{\beta}{2}-\sin\frac{\alpha}{2}\sin\frac{\beta}{2}}=-\frac{\cos\frac{\gamma}{2}\cos\frac{\delta}{2}+\sin\frac{\gamma}{2}\sin\frac{\delta}{2}}{\cos\frac{\gamma}{2}\cos\frac{\delta}{2}-\sin\frac{\gamma}{2}\sin\frac{\delta}{2}}$
Now apply componendo and dividendo on both sides,we will get the answer.
Since it's lack of the contexts for the cases of ellipse and parabola, I'd like to start from a circle:
$$(x,y)=r(\cos \theta,\sin \theta)$$
Equation of chord $AB$
$$x\cos \frac{\alpha+\beta}{2}+y\sin \frac{\alpha+\beta}{2} =r\cos \frac{\alpha-\beta}{2}$$
$$m_1=-\cot \frac{\alpha+\beta}{2}$$
From properties of cyclic quadrilaterals:
$$180^{\circ}=A+C=B+D$$
$$0=\tan B+\tan D$$
$$0=\frac{m_1-m_2}{1+m_1 m_2}+\frac{m_3-m_4}{1+m_3 m_4}$$
which is equivalent to
$$0=\tan \frac{\alpha-\gamma}{2}+\tan \frac{\gamma-\alpha}{2}$$
$$m_1-m_2+m_3-m_4=m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3$$
For an ellipse $(x,y)=(a\cos \theta,b\sin \theta)$
$$m_1=-\frac{b}{a}\cot \frac{\alpha+\beta}{2}$$
Transform every $m_i$ back to the auxiliary circle,
$$\frac{a}{b}(m_1-m_2+m_3-m_4)= \frac{a^3}{b^3}(m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3)$$
$$\frac{m_1-m_2+m_3-m_4}{m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3} =\frac{a^2}{b^2}$$
For a hyperbola $(x,y)=(a\cosh t,b\sinh t)=(a\cos it,-bi\sin it)$
$$m_1=\frac{b}{a}\coth \frac{t+u}{2} =\frac{bi}{a} \cot \frac{i(t+u)}{2}$$
$\fbox{Note that $t, u, v, w$ differ from the eccentricity angles $\alpha, \beta, \gamma, \delta$ in $(a\sec , b\tan )$.}$
Now replace $b$ by $bi$, we have
$$\frac{m_1-m_2+m_3-m_4}{m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3} =\frac{a^2}{(bi)^2}$$
$$\frac{m_1-m_2+m_3-m_4}{m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3} =-\frac{a^2}{b^2}$$
For a parabola of the form $x^2=4py$, take $b\to \infty$
$$m_1-m_2+m_3-m_4=0$$
For a parabola of the form $y^2=4qx$, take $a\to \infty$.
$$m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3=0$$
Best Answer
WLOG the equations of the Circle & the Ellipse can be chosen to be $$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\text{ and } x^2+y^2=r^2$$
We know the parametric coordinate of any $P(x,y)$ on the ellipse is $(h+a\cos\phi,k+b\sin\phi)$ where $\phi$ is the eccentric angle
To find the intersection, $$(h+a\cos\phi)^2+(k+b\sin\phi)^2=r^2$$
Use Weierstrass substitution, to form a Bi-Quadratic Equation in $\tan\frac\phi2$ whose roots are $\displaystyle\tan\frac\alpha2,\tan\frac\beta2,\tan\frac\gamma2,\tan\frac\delta2$
Now using $\tan(A+B+C+D)$ formula ,
$$t^4\{(h-a)^2+k^2-r^2\}+(4bk)t^3+()t^2+(4bk)t+()=0$$
Using Vieta's formula $$\sum \tan\frac\alpha2=-\frac{4bk}{(h-a)^2+k^2-r^2}=\sum \tan\frac\alpha2\tan\frac\beta2\tan\frac\gamma2$$
$$\implies\tan\left(\frac{\alpha+\beta+\gamma+\delta}2\right)=0\implies \frac{\alpha+\beta+\gamma+\delta}2=n\pi$$ where $n$ is any integer