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

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(B) Let the two points on the ellipse be $A(a \cos 	heta_1, b \sin 	heta_1)$ and $B(a \cos 	heta_2, b \sin 	heta_2)$.Let the center of the ellipse

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(B) Let the two points on the ellipse be $A(a \cos 	heta_1, b \sin 	heta_1)$ and $B(a \cos 	heta_2, b \sin 	heta_2)$.Let the center of the ellipse be $O(0, 0)$.The slope of $OA$ is $m_1 = \frac{b \sin 	heta_1}{a \cos 	heta_1} = \frac{b}{a} 	an 	heta_1$.The slope of $OB$ is $m_2 = \frac{b \sin 	heta_2}{a \cos 	heta_2} = \frac{b}{a} 	an 	heta_2$.For the chord $AB$ to subtend a right angle at the center $O$,the product of the slopes $m_1$ and $m_2$ must be $-1$.$m_1 	imes m_2 = \left(\frac{b}{a} 	an 	heta_1ight) 	imes \left(\frac{b}{a} 	an 	heta_2ight) = \frac{b^2}{a^2} (	an 	heta_1 	an 	heta_2)$.Given $	an 	heta_1 	an 	heta_2 = -\frac{a^2}{b^2}$,we substitute this into the product:$m_1 	imes m_2 = \frac{b^2}{a^2} 	imes \left(-\frac{a^2}{b^2}ight) = -1$.Since the product of the slopes is $-1$,the lines $OA$ and $OB$ are perpendicular.Thus,the chord $AB$ subtends a right angle at the center.

If $\tan \theta_1 \times \tan \theta_2 = -\frac{a^2}{b^2}$,then the chord joining $2$ 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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