Locus of centroid of triangle formed by any point P on hyperbola $16 x^{2} - 9 y^{2} + 32 x + 36 y - 164 = 0$ and it’s foci is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

$9 x^{2} - 16 y^{2} + 36 x + 32 y - 144 = 0$

$16 x^{2} - 9 y^{2} + 32 x + 36 y - 144 = 0$

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

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} 16 x^{2} + 32 x - 9 y^{2} + 36 y = 164 \\ 16 {(x + 1)}^{2} - 9 {(y - 2)}^{2} = 164 + 16 - 36 = 144 \\ \frac{{(x + 1)}^{2}}{9} - \frac{{(y - 2)}^{2}}{16} = 1 \\ e = \sqrt{1 + \frac{b^{2}}{a^{2}}} \Rightarrow \sqrt{1 + \frac{16}{9}} = \frac{5}{3} \end{array}$
Foci are $(- 1 \pm a e, 0 + 2) \Rightarrow (4, 2) a n d (- 6, 2)$
Question Image
Let centroid of $Δ A B C$ be $(h, k) & A (α, β)$ be point on hyperbola
$\begin{array}{l} ∴ h = \frac{α - 6 + 4}{3}, k = \frac{β + 2 + 2}{3} \\ α = 3 h + 2, β = 3 k - 4 \end{array}$
Since $(α, β)$ lies on hyperbola
$\begin{array}{l} ∴ 16 {(3 h + 2 + 1)}^{2} - 9 {(3 k - 4 - 2)}^{2} = 144 \\ 144 {(h + 1)}^{2} - 81 {(k - 2)}^{2} = 144 \\ 16 (h^{2} + 2 h + 1) - 9 (k^{2} - 4 k + 4) = 16 \\ 16 x^{2} - 9 y^{2} + 32 x + 36 y - 36 = 0 \end{array}$