Question

If $\theta$and $\varphi$are eccentric angles of the ends of a pair of conjugate diameters of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,then $\theta - \varphi$is equal to

• A. $\pm \frac{\pi}{2}$
• B. $\pm \pi$
• C. 0
• D. None of these

Answer 1

Answer: (A)

$\pm \frac{\pi}{2}$

Answer 2

Let $y = m_{1}x$and $y = m_{2}x$be a pair of conjugate diameter of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$and let $P(a\cos\theta,b\sin\theta)$ and $Q(a\cos\varphi,b\sin\varphi)$ be ends of these two diameters. Then $m_{1}m_{2} = \frac{- b^{2}}{a^{2}}$

⇒$\frac{b\sin\theta - 0}{a\cos\theta - 0} \times \frac{b\sin\varphi - 0}{a\cos\varphi - 0} = \frac{- b^{2}}{a^{2}}$⇒$\sin\theta\sin\varphi = - \cos\theta\cos\varphi$

⇒ $\cos(\theta - \varphi) = 0$⇒ $\theta - \varphi = \pm \pi/2$