Question

The area of the triangle formed by three points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ whose eccentric angles are $\alpha ,\beta$ and $\gamma$ is:

• A. $ab\sin \frac{\alpha -\beta }{2}\cos \frac{\beta -\gamma }{2}\cos \frac{\gamma -\alpha }{2}$
• B. $ab\sin \frac{\alpha -\beta }{2}\sin \frac{\beta -\gamma }{2}\cos \frac{\gamma -\alpha }{2}$
• C. $ab\sin \frac{\alpha -\beta }{2}\sin \frac{\beta -\gamma }{2}\sin \frac{\gamma -\alpha }{2}$
• D. $ab\cos \frac{\alpha -\beta }{2}\cos \frac{\beta -\gamma }{2}\cos \frac{\gamma -\alpha }{2}$

Answer 1

The correct option is C $ab\sin \frac{\alpha -\beta }{2}\sin \frac{\beta -\gamma }{2}\sin \frac{\gamma -\alpha }{2}$

$△=\frac{1}{2}\left|\begin{array}{ccc}1& a\cos \alpha & b\sin \alpha \\ 1& a\cos \beta & b\sin \beta \\ 1& a\cos \gamma & b\sin \gamma \end{array}\right|$

$R_2\to R_2-R_1,R_3\to R_3-R_1$

$△=\frac{1}{2}ab\left|\begin{array}{ccc}1& a\cos \alpha & b\sin \alpha \\ 0& \cos \beta -\cos \alpha & \sin \beta -\sin \alpha \\ 0& \cos \gamma -\cos \alpha & \sin \gamma -\sin \alpha \end{array}\right|$

$△=\frac{1}{2}ab\left|\begin{array}{ccc}1& a\cos \alpha & b\sin \alpha \\ 0& 2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}& 2\sin \frac{\beta -\alpha }{2}\cos \frac{\beta +\alpha }{2}\\ 0& -2\sin \frac{\alpha +\gamma }{2}\sin \frac{\gamma -\alpha }{2}& 2\sin \frac{\gamma -\alpha }{2}\cos \frac{\gamma +\alpha }{2}\end{array}\right|$

$=ab\sin \frac{\beta -\gamma }{2}\sin \frac{\alpha -\beta }{2}\sin \frac{\gamma -\alpha }{2}$