If tan theta_1 cdot tan theta_2 = -frac{a^2}{ | vedclass

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(B) Let the points on the ellipse be $P(a \cos 	heta_1, b \sin 	heta_1)$ and $Q(a \cos 	heta_2, b \sin 	heta_2)$.The slopes of the lines joining t

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(B) Let the points on the ellipse be $P(a \cos 	heta_1, b \sin 	heta_1)$ and $Q(a \cos 	heta_2, b \sin 	heta_2)$.The slopes of the lines joining the origin $O(0,0)$ to $P$ and $Q$ are $m_1 = \frac{b \sin 	heta_1}{a \cos 	heta_1} = \frac{b}{a} 	an 	heta_1$ and $m_2 = \frac{b \sin 	heta_2}{a \cos 	heta_2} = \frac{b}{a} 	an 	heta_2$.The product of the slopes is $m_1 m_2 = \frac{b^2}{a^2} 	an 	heta_1 	an 	heta_2$.Given $	an 	heta_1 	an 	heta_2 = -\frac{a^2}{b^2}$,we have $m_1 m_2 = \frac{b^2}{a^2} \left(-\frac{a^2}{b^2}ight) = -1$.Since the product of the slopes is $-1$,the lines $OP$ and $OQ$ are perpendicular,meaning the chord $PQ$ subtends a right angle at the centre $O(0,0)$.

If $\tan \theta_1 \cdot \tan \theta_2 = -\frac{a^2}{b^2}$,then the chord joining two points $\theta_1$ and $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ will subtend a right angle at:

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