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

Question

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

Dr. Abubacker · Accepted Answer

(a)
Let $y = m_{1}x$ and $y = m_{2}x$ be a pair of conjugate diameters of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and let $P(a\cos{	heta,\ {b\ sin}{	heta)}}$ and $Q(a\cos{\phi,\ b}\sin{\phi)}$ be ends of these two diameters. Then,
$m_{1}m_{2} = - \frac{b^{2}}{a^{2}}$
$\Rightarrow \frac{{b\ sin}	heta - 0}{a\cos	heta - 0} 	imes \frac{{b\ sin}{\phi - 0}}{a\cos\phi - 0} = - \frac{b^{2}}{a^{2}}$
$\Rightarrow \sin	heta\sin\phi = - \cos	heta\cos\phi$
$\Rightarrow \cos{(	heta} - \phi) = 0$
$\Rightarrow 	heta - \phi = \pm \frac{\pi}{2}$

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

Explanation

Reviewed by Dr. Abubacker

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