Question

Prove that the area of the triangle formed by three points on an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, whose eccentric angles are $\theta, \phi$, and $\psi$, is $|2ab \sin \frac{\phi - \psi}{2} \sin \frac{\psi - \theta}{2} \sin \frac{\theta - \phi}{2}|$.

Answer 1

The area of the triangle formed by three points on an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, whose eccentric angles are $\theta, \phi$, and $\psi$, is $|2ab \sin \frac{\phi - \psi}{2} \sin \frac{\psi - \theta}{2} \sin \frac{\theta - \phi}{2}|$.

Answer 2

The coordinates of the three points on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with eccentric angles $\theta, \phi, \psi$ are $(a \cos \theta, b \sin \theta)$, $(a \cos \phi, b \sin \phi)$, and $(a \cos \psi, b \sin \psi)$. The area of the triangle formed by these points is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Substituting the coordinates and simplifying yields $\frac{ab}{2} |\sin(\phi - \theta) + \sin(\psi - \phi) + \sin(\theta - \psi)|$. Using the identity $\sin A + \sin B + \sin C = -4 \sin(A/2) \sin(B/2) \sin(C/2)$ when $A+B+C=0$, the area becomes $\frac{ab}{2} |-4 \sin(\frac{\phi - \theta}{2}) \sin(\frac{\psi - \phi}{2}) \sin(\frac{\theta - \psi}{2})|$. This simplifies to $2ab |\sin(\frac{\phi - \theta}{2}) \sin(\frac{\psi - \phi}{2}) \sin(\frac{\theta - \psi}{2})|$, which is equal to $|2ab \sin \frac{\phi - \psi}{2} \sin \frac{\psi - \theta}{2} \sin \frac{\theta - \phi}{2}|$.