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JEE Main202125 Jul 2021Morning ShiftMathematicsHyperbolaActual

The locus of the centroid of the triangle formed by any point P on the hyperbola 16 x 2 - 9 y 2 + 32 x + 36 y - 164 = 0 and its foci is

Options

  1. A16 x 2 - 9 y 2 + 32 x + 36 y - 36 = 0
  2. B9 x 2 - 16 y 2 + 36 x + 32 y - 144 = 0
  3. C16 x 2 - 9 y 2 + 32 x + 36 y - 144 = 0
  4. D9 x 2 - 16 y 2 + 36 x + 32 y - 36 = 0

Correct answer

A. 16 x 2 - 9 y 2 + 32 x + 36 y - 36 = 0

Step-by-step solution

Given hyperbola is 16 x 2 - 9 y 2 + 32 x + 36 y - 164 = 0 ⇒ 16 x + 1 2 - 9 y - 2 2 = 164 + 16 - 36 ⇒ 16 x + 1 2 - 9 y - 2 2 = 144 . . . . 1 ⇒ ( x + 1 ) 2 9 - ( y - 2 ) 2 16 = 1 Compare the above equation with ⇒ ( x - h ) 2 a 2 - ( y - k ) 2 b 2 = 1 We get, h = - 1 ,   k =   2 ,   a 2 = 9 and b 2 = 16 ∴ Eccentricity, e = 1 + 16 9 = 5 3 We know that, focii are h + a e ,   k and h - a e ,   k Hence, focii are ( 4 , 2 ) and ( - 6 , 2 ) Let the centroid be ( h , k )

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