Question

If Normals at Pts $\alpha, \beta, \gamma, \delta$ on the Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are congruent, then find $\sum \cos\alpha \sum \sec\alpha=?$

• A. 0
• B. 4
• C. 16
• D. 1

Answer 1

Answer: (A)

0

Answer 2

The length of the normal segment from a point $(a\cos\theta, b\sin\theta)$ on the ellipse to the x-axis is given by $L_x(\theta) = b\sqrt{1 - \frac{a^2-b^2}{a^2}\cos^2\theta}$. If the normals at $\alpha, \beta, \gamma, \delta$ are congruent, then $L_x(\alpha) = L_x(\beta) = L_x(\gamma) = L_x(\delta)$. This implies that $\cos^2\alpha = \cos^2\beta = \cos^2\gamma = \cos^2\delta$. Let this common value be $C$. Then $\cos\theta = \pm \sqrt{C}$. For four distinct points, the set of cosine values must be $\{\cos\alpha, \cos\beta, \cos\gamma, \cos\delta\} = \{c, c, -c, -c\}$, where $c = \sqrt{C}$. Then, $\sum \cos\alpha = c + c + (-c) + (-c) = 0$. And $\sum \sec\alpha = \frac{1}{c} + \frac{1}{c} + \frac{1}{-c} + \frac{1}{-c} = 0$. Therefore, $\sum \cos\alpha \sum \sec\alpha = (0)(0) = 0$.