The chord joining two points theta_1 and thet | vedclass

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(B) Let $P(a \cos 	heta_1, b \sin 	heta_1)$ and $Q(a \cos 	heta_2, b \sin 	heta_2)$ be two points on the ellipse.Let $O(0, 0)$ be the center of th

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

The chord joining two points $\theta_1$ and $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ subtends a right angle at the . . . point. (Given $\tan \theta_1 \tan \theta_2 = -\frac{a^2}{b^2}$)

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