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A chord is drawn passing through P(2,2) ...

A chord is drawn passing through `P(2,2)` on the ellipse `(x^(2))/(25)+(y^(2))/(16) =1` such that it intersects the ellipse at A and B. Then maximum value of `PA.PB` is

A

`(61)/(4)`

B

`(59)/(4)`

C

`(71)/(4)`

D

`(63)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
B
Let equation of secant be
`(x-2)/(cos theta) = (y-2)/(sin theta) =r` (Parametric form)
Solving it with ellipse
`((r cos theta+2)^(2))/(25)+((r sin theta +2)^(2))/(16)=1`
`r^(2) (16 cos^(2) theta + 25 sin^(2) theta) +r (64 cos theta + 100 sin theta) - 236=0`
`|r_(1)r_(2)| = PA. PB = |(-236)/(16 cos^(2)theta+25 sin^(2)theta)|`
`=|-(236)/(16+9 sin^(2)theta)|`
Max. `PA. PB = (236)/(16) = (59)/(4)`
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