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QJEEM21Ma3873: JEE Main 2021 Mathematics - Class 12 JEEM21Ma3873 Solution: The locus of the centroid of the triangle formed by any point P …

The locus of the centroid of the triangle formed by any point P on the hyperbola 16x2−9y2+32x+36y−164=016{x^2} - 9{y^2} + 32x + 36y - 164 = 0 , and its foci is :

Multiple Choice Options

  • A 16x2−9y2+32x+36y−36=016{x^2} - 9{y^2} + 32x + 36y - 36 = 0
  • B 9x2−16y2+36x+32y−144=09{x^2} - 16{y^2} + 36x + 32y - 144 = 0
  • C 16x2−9y2+32x+36y−144=016{x^2} - 9{y^2} + 32x + 36y - 144 = 0
  • D 9x2−16y2+36x+32y−36=09{x^2} - 16{y^2} + 36x + 32y - 36 = 0

Step-by-Step Verified Solution

Given hyperbola is

16(x+1)2−9(y−2)2=164+16−36=14416{(x + 1)^2} - 9{(y - 2)^2} = 164 + 16 - 36 = 144

⇒(x+1)29−(y−2)216=1\Rightarrow {{{{(x + 1)}^2}} \over 9} - {{{{(y - 2)}^2}} \over {16}} = 1

Eccentricity, e=1+169=53e = \sqrt {1 + {{16} \over 9}} = {5 \over 3}

⇒\Rightarrow foci are (4, 2) and ( −- 6, 2)

JEE Main 2021 (Online) 25th July Morning Shift Mathematics - Hyperbola Question 41 English Explanation
Let the centroid be (h, k) & A( α\alpha , β\beta ) be point on hyperbola.

So, h=α−6+43,k=β+2+23h = {{\alpha - 6 + 4} \over 3},k = {{\beta + 2 + 2} \over 3}

⇒α=3h+2,β=3k−4\Rightarrow \alpha = 3h + 2,\beta = 3k - 4

( α\alpha , β\beta ) lies on hyperbola so

16(3h+2+1)2−9(3k−4−2)2=14416{(3h + 2 + 1)^2} - 9{(3k - 4 - 2)^2} = 144

⇒144(h+1)2−81(k−2)2=144\Rightarrow 144{(h + 1)^2} - 81{(k - 2)^2} = 144

⇒16(h2+2h+1)−9(k2−4k+4)=16\Rightarrow 16({h^2} + 2h + 1) - 9({k^2} - 4k + 4) = 16

⇒16x2−9y2+32x+36y−36=0\Rightarrow 16{x^2} - 9{y^2} + 32x + 36y - 36 = 0

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