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Sum of Focal Radii Expression Equals 2a | JEE

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

Question
If b2=a2(1−e2)b^2=a^2(1-e^2), then for any α\alpha, a,b∈Ra,b\in\mathbb{R},
a2cos⁡2α+b2sin⁡2α−2a2ecos⁡α+a2e2+a2cos⁡2α+b2sin⁡2α+2a2ecos⁡α+a2e2\sqrt{a^2\cos^2\alpha+b^2\sin^2\alpha-2a^2e\cos\alpha+a^2e^2}+\sqrt{a^2\cos^2\alpha+b^2\sin^2\alpha+2a^2e\cos\alpha+a^2e^2}
 is equal to:
Aa+ba+b
Baa
C2a2acorrect
D2b2b
Solution
Step 1: Recognise the point and foci. Let P=(acos⁡α, bsin⁡α)P=(a\cos\alpha,\ b\sin\alpha) lie on x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1, with foci S=(ae,0)S=(ae,0) and S′=(−ae,0)S'=(-ae,0). Step 2: Compute the focal radius SPSP.
SP=(acos⁡α−ae)2+(bsin⁡α)2=a2cos⁡2α−2a2ecos⁡α+a2e2+b2sin⁡2α,SP=\sqrt{(a\cos\alpha-ae)^2+(b\sin\alpha)^2}=\sqrt{a^2\cos^2\alpha-2a^2e\cos\alpha+a^2e^2+b^2\sin^2\alpha},
 which is the first radical. Step 3: Similarly S′P=a2cos⁡2α+2a2ecos⁡α+a2e2+b2sin⁡2αS'P=\sqrt{a^2\cos^2\alpha+2a^2e\cos\alpha+a^2e^2+b^2\sin^2\alpha}, the second radical. Step 4: By the defining property of the ellipse, SP+S′P=2aSP+S'P=2a. Correct answer: (3)
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