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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

S

meets the ellipse

9x^2+16y^2=144

P

and

Q

A

and

B

are the feet of the perpendiculars from the foci

S

and

S'

to the tangent at

P

, and

O

is the centre. Then

\dfrac{SP+SQ}{SP\cdot SQ}+\dfrac{OA\cdot OB}{SA\cdot S'B}

equals:

\dfrac{9}{24}

\dfrac{25}{9}

\dfrac{24}{9}

correct

\dfrac{9}{16}

Solution

Write

9x^2+16y^2=144

\dfrac{x^2}{16}+\dfrac{y^2}{9}=1

, so

a=4,\ b=3

. Step 1: For a focal chord, the semi-latus rectum

\ell=\dfrac{b^2}{a}=\dfrac94

is the harmonic mean of the segments:

\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=a

, giving

OA\cdot OB=a^2=16

. Also the product of perpendiculars from the two foci to a tangent is

b^2

, so

SA\cdot S'B=b^2=9

. Hence

\frac{OA\cdot OB}{SA\cdot S'B}=\frac{a^2}{b^2}=\frac{16}{9}.

Step 3: Add.

\frac{8}{9}+\frac{16}{9}=\frac{24}{9}.

Correct answer: (3)

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