The locus of the centroid of the triangle formed by any point P on the hyperbola 16 x² - 9 y² + 32 x + 36 y - 164 = 0, and its foci is : [25 Jul 2021 Shift 1]

Correct Answer is : 16x2−9y2+32x+36y−36=016 x² - 9 y² + 32 x + 36 y - 36 = 016x2−9y2+32x+36y−36=0

16 x² - 9 y² + 32 x + 36 y - 36 = 0

9 x² - 16 y² + 36 x + 32 y - 144 = 0

16 x² - 9 y² + 32 x + 36 y - 144 = 0

9 x² - 16 y² + 36 x + 32 y - 36 = 0

Correct Answer is : 16x2−9y2+32x+36y−36=016 x² - 9 y² + 32 x + 36 y - 36 = 016x2−9y2+32x+36y−36=0

Test Index

Hyperbola

Section: Mathematics

Question : 61 of 98

Marks: +1, -0

P

16 x^{2} - 9 y^{2} + 32 x + 36 y - 164 = 0

Solution:

Given hyperbola is

16(x+1)^{2} - 9(y-2)^{2}

=164+16-36=144

\Rightarrow \frac{(x+1)^{2}}{9} - \frac{(y-2)^{2}}{16} = 1

Eccentricity,

e = \sqrt{1 + \frac{16}{9}} = \frac{5}{3}

\Rightarrow

foci are

(4,2)

and

(-6,2)

Let the centroid be

(h, k)

A(\alpha, \beta)

be point on hyperbola

h = \frac{\alpha - 6 + 4}{3}, k = \frac{\beta + 2 + 2}{3}

\Rightarrow \alpha = 3 h + 2, \beta = 3 k - 4

(\alpha, \beta)

lies on hyperbola so

16(3 h+2+1)^{2} - 9(3 k-4-2)^{2}

=144

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

\Rightarrow 16(h^{2}+2 h+1) - 9(k^{2} - 4 k

+4)=16

\Rightarrow 16 x^{2} - 9 y^{2} + 32 x

+36 y - 36 = 0

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