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Chord Subtending Right Angle at the Centre of an Ellipse | JEE

JEE Maths question with a full step-by-step solution.

Question
If tan⁡θ1tan⁡θ2=−a2b2\tan\theta_1\tan\theta_2=-\dfrac{a^2}{b^2}, then the chord joining the points θ1\theta_1 and θ2\theta_2 on the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 subtends a right angle at:
Afocus
Bcentrecorrect
Cend of the major axis
Dend of the minor axis
Solution
Solution: Step 1: Let P=(acos⁡θ1,bsin⁡θ1)P=(a\cos\theta_1,b\sin\theta_1) and Q=(acos⁡θ2,bsin⁡θ2)Q=(a\cos\theta_2,b\sin\theta_2), with OO the centre. Step 2: Slopes of OPOP and OQOQ:
m1=bsin⁡θ1acos⁡θ1=batan⁡θ1,m2=batan⁡θ2.m_1=\frac{b\sin\theta_1}{a\cos\theta_1}=\frac{b}{a}\tan\theta_1,\qquad m_2=\frac{b}{a}\tan\theta_2.
 Step 3: Their product is
m1m2=b2a2tan⁡θ1tan⁡θ2=b2a2(−a2b2)=−1.m_1m_2=\frac{b^2}{a^2}\tan\theta_1\tan\theta_2=\frac{b^2}{a^2}\left(-\frac{a^2}{b^2}\right)=-1.
 Step 4: Since OP⊥OQOP\perp OQ, the chord subtends a right angle at the centre. Correct answer: (2)
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