A line passing through the point P( 5, 5) intersects the ellipse x236 + y225 = 1 at A and B such that (PA) (PB) is maximum. Then 5 (PA2 + PB2 ) is equal to :
- 290
- 377
- 338
- 218
Answer
(3) 338
Solution
Any point on the line: ( 5 + r , 5 + r ). Putting it in the ellipse:
r2 ( 2 36 + 2 25 ) + 2 5 r ( 36 + 25 ) + ( 536 + 525 - 1 ) = 0
∵ P lies inside the ellipse, PA PB = |r1 r2| = 119/180 2 36 + 2 25
This is maximum when the denominator is least, i.e. = 0 (denominator = 136) ⇒ the line is y = 5.
Then x236 = 1 - 525 = 45 ⇒ x = 12 5
⇒ PA = 12 5 - 5 = 7 5, PB = 12 5 + 5 = 17 5
∴ 5 (PA2 + PB2 ) = 5 ( 495 + 2895 ) = 338
Answer: (3)