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Question: Answered & Verified by Expert
A point moves so that the sum of its distances from (a e, 0) and (-a e, 0) is 2 a, then the equaton to its locus where b2=a21-e2 is
MathematicsStraight LinesAP EAMCETAP EAMCET 2021 (20 Aug Shift 2)
Options:
  • A x2a2-y2b2=1
  • B x2a2+y2b2=1
  • C x2b2+y2a2=1
  • D y2b2-x2a2=1
Solution:
1800 Upvotes Verified Answer
The correct answer is: x2a2+y2b2=1

Suppose that the co-ordinates of the moving point be x,y, then from the given condition

x-ae2+y2+x+ae2+y2=2a  .....1

Now

x-ae2+y2-x+ae2+y2=-4aex  ....2

 a-b2-a+b2=-4ab

On dividing 2 by 1, we get

x-ae2+y2-x+ae2+y2x-ae2+y2+x+ae2+y2=-4aex2a

x-ae2+y22-x+ae2+y22x-ae2+y2+x+ae2+y2=-2ex

x-ae2+y2-x+ae2+y2=-2ex......3

Adding 1 and 3 we get

2x-ae2+y2=2a-ex.

Squaring on both sides, we get

x2-2aex+a2e2+y2=a2-2aex+e2x2

 x21-e2+y2=a21-e2

x2a2+y2a21-e2=1

x2a2+y2b2=1

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