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(A) Let $y = m_1x$ and $y = m_2x$ be a pair of conjugate diameters of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Let $P(a \cos 	heta, b \sin

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

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(A) Let $y = m_1x$ and $y = m_2x$ be a pair of conjugate diameters of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Let $P(a \cos 	heta, b \sin 	heta)$ and $Q(a \cos \phi, b \sin \phi)$ be the ends of these two diameters.The condition for conjugate diameters is $m_1 m_2 = -\frac{b^2}{a^2}$.Since $m_1 = \frac{b \sin 	heta}{a \cos 	heta}$ and $m_2 = \frac{b \sin \phi}{a \cos \phi}$,we have:$\left(\frac{b \sin 	heta}{a \cos 	heta}ight) 	imes \left(\frac{b \sin \phi}{a \cos \phi}ight) = -\frac{b^2}{a^2}$.This simplifies to $\frac{\sin 	heta \sin \phi}{\cos 	heta \cos \phi} = -1$.Therefore,$\sin 	heta \sin \phi + \cos 	heta \cos \phi = 0$,which is $\cos(	heta - \phi) = 0$.Thus,$	heta - \phi = \pm \frac{\pi}{2}$.

If $\theta$ and $\phi$ 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 $\theta - \phi$ is equal to

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