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Focal Chord and Tangent Perpendicular Identity | JEE

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

Question
A focal chord through the focus SS meets the ellipse 9x2+16y2=1449x^2+16y^2=144 at PP and QQ. AA and BB are the feet of the perpendiculars from the foci SS and S′S' to the tangent at PP, and OO is the centre. Then SP+SQSP⋅SQ+OA⋅OBSA⋅S′B\dfrac{SP+SQ}{SP\cdot SQ}+\dfrac{OA\cdot OB}{SA\cdot S'B} equals:
A924\dfrac{9}{24}
B259\dfrac{25}{9}
C249\dfrac{24}{9}correct
D916\dfrac{9}{16}
Solution
Write 9x2+16y2=1449x^2+16y^2=144 as x216+y29=1\dfrac{x^2}{16}+\dfrac{y^2}{9}=1, so a=4, b=3a=4,\ b=3. Step 1: For a focal chord, the semi-latus rectum ℓ=b2a=94\ell=\dfrac{b^2}{a}=\dfrac94 is the harmonic mean of the segments:
1SP+1SQ=2ℓ=2ab2=89  ⇒  SP+SQSP⋅SQ=89.\frac{1}{SP}+\frac{1}{SQ}=\frac{2}{\ell}=\frac{2a}{b^2}=\frac{8}{9}\;\Rightarrow\; \frac{SP+SQ}{SP\cdot SQ}=\frac{8}{9}.
 Step 2: The feet of the perpendiculars from the foci to a tangent lie on the auxiliary circle, so OA=OB=aOA=OB=a, giving OA⋅OB=a2=16OA\cdot OB=a^2=16. Also the product of perpendiculars from the two foci to a tangent is b2b^2, so SA⋅S′B=b2=9SA\cdot S'B=b^2=9. Hence
OA⋅OBSA⋅S′B=a2b2=169.\frac{OA\cdot OB}{SA\cdot S'B}=\frac{a^2}{b^2}=\frac{16}{9}.
 Step 3: Add.
89+169=249.\frac{8}{9}+\frac{16}{9}=\frac{24}{9}.
 Correct answer: (3)
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