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JEE Main20198 Apr 2019Morning ShiftMathematicsEllipseActual

Let O 0,0 and A 0,1 be two fixed points. Then, the locus of a point P such that the perimeter of Δ A O P is 4 is

Options

  1. A8 x 2 + 9 y 2 - 9 y = 18
  2. B9 x 2 - 8 y 2 + 8 y = 16
  3. C8 x 2 - 9 y 2 + 9 y = 18
  4. D9 x 2 + 8 y 2 - 8 y = 16

Correct answer

D. 9 x 2 + 8 y 2 - 8 y = 16

Step-by-step solution

Given, O A + P O + P A = 4   ⇒ P O + P A = 3   ⇒ Locus of P is ellipse with foci at O & A and major axis 2 b = 3 Distance between foci = 2 b e = 1 ⇒ e = 1 3 ⇒ Minor axis 2 a = 2 b 1 - e 2 = 3 . 2 2 3 = 2 2 ⇒ Equation of Locus of P is : x 2 8 + y - 1 2 2 9 = 1 4 By simplifying the above equation, we get 9 x 2 + 8 y 2 - 8 y = 16

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