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Chord Condition: Eccentric Angles Differ by pi/2 | JEE

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

Question
The line lx+my+n=0lx+my+n=0 cuts the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 in points whose eccentric angles differ by π2\dfrac{\pi}{2}, if:
Aa2l2+b2n2=2m2a^2l^2+b^2n^2=2m^2
Ba2m2+b2l2=2n2a^2m^2+b^2l^2=2n^2
Ca2l2+b2m2=2n2a^2l^2+b^2m^2=2n^2correct
Da2n2+b2m2=2l2a^2n^2+b^2m^2=2l^2
Solution
Step 1: The chord joining eccentric angles α\alpha and β\beta is
xacos⁡α+β2+ybsin⁡α+β2=cos⁡α−β2.\frac{x}{a}\cos\frac{\alpha+\beta}{2}+\frac{y}{b}\sin\frac{\alpha+\beta}{2}=\cos\frac{\alpha-\beta}{2}.
 Step 2: Put β=α+π2\beta=\alpha+\dfrac{\pi}{2}, so α−β2=−π4\dfrac{\alpha-\beta}{2}=-\dfrac{\pi}{4} and cos⁡α−β2=12\cos\dfrac{\alpha-\beta}{2}=\dfrac{1}{\sqrt2}. After clearing 2\sqrt2 the chord reduces to
bx(cos⁡α−sin⁡α)+ay(cos⁡α+sin⁡α)=ab.bx(\cos\alpha-\sin\alpha)+ay(\cos\alpha+\sin\alpha)=ab.
 Step 3: Compare with lx+my=−nlx+my=-n. Matching coefficients,
cos⁡α−sin⁡α=−aln,cos⁡α+sin⁡α=−bmn.\cos\alpha-\sin\alpha=-\frac{al}{n},\qquad \cos\alpha+\sin\alpha=-\frac{bm}{n}.
 Step 4: Use the identity (cos⁡α−sin⁡α)2+(cos⁡α+sin⁡α)2=2(\cos\alpha-\sin\alpha)^2+(\cos\alpha+\sin\alpha)^2=2.
a2l2n2+b2m2n2=2  ⇒  a2l2+b2m2=2n2.\frac{a^2l^2}{n^2}+\frac{b^2m^2}{n^2}=2 \;\Rightarrow\; a^2l^2+b^2m^2=2n^2.
 Correct answer: (3)
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