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AP EAMCET202120 Aug 2021Evening ShiftMathematicsStraight LinesActual

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 b 2 = a 2 1 - e 2 is

Options

  1. Ax 2 a 2 - y 2 b 2 = 1
  2. Bx 2 a 2 + y 2 b 2 = 1
  3. Cx 2 b 2 + y 2 a 2 = 1
  4. Dy 2 b 2 - x 2 a 2 = 1

Correct answer

B. x 2 a 2 + y 2 b 2 = 1

Step-by-step solution

Suppose that the co-ordinates of the moving point be x , y ,   then from the given condition x - a e 2 + y 2 + x + a e 2 + y 2 = 2 a     . . . . . 1 Now x - a e 2 + y 2 - x + a e 2 + y 2 = - 4 a e x     . . . . 2 ∵   a - b 2 - a + b 2 = - 4 a b On dividing 2   by 1 , we get x - a e 2 + y 2 - x + a e 2 + y 2 x - a e 2 + y 2 + x + a e 2 + y 2 = - 4 a e x 2 a ⇒ x - a e 2 + y 2 2 - x + a e 2 + y 2 2 x - a e 2 + y 2 + x + a e 2 + y 2 = - 2 e x ⇒ x - a e 2 + y 2 - x +

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