The locus of the centroid of the triangle formed by any point P on...

Given hyperbola is

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

$⇒ 16{\left(x + 1\right)}^{2} - 9{\left(y - 2\right)}^{2} = 164 + 16 - 36$

$⇒ 16{\left(x + 1\right)}^{2} - 9{\left(y - 2\right)}^{2} = 144 . . . . \left(1\right)$

$⇒ \dfrac{( x + 1{)}^{2}}{9} - \dfrac{( y - 2{)}^{2}}{16} = 1$

Compare the above equation with $⇒ \dfrac{( x - h{)}^{2}}{{a}^{2}} - \dfrac{( y - k{)}^{2}}{{b}^{2}} = 1$

We get, $h = - 1, k = 2, {a}^{2} = 9 and {b}^{2} = 16$

$∴$ Eccentricity, $e = \sqrt{1 + \dfrac{16}{9}} = \dfrac{5}{3}$

We know that, focii are $\left(h + ae, k\right)$ and \left(h - ae, k\right)$</p><p>Hence, focii are$( 4, 2 ) and ( - 6, 2 )$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/828dd66d-227a-4e1b-b8ac-5aa60cb0406d-image.png" style="width: 253px; height: 178px;" /></p><p>Let the centroid be$( h, k )&A ( \alpha, \beta )$be point on hyperbola</p><p>So$h = \dfrac{\alpha - 6 + 4}{3}, k = \dfrac{\beta + 2 + 2}{3}$</p><p>$⇒ \alpha = 3h + 2, \beta = 3k - 4$</p><p>Put the values of$\alpha, \beta$in the equation\left(1\right)$,

$16{\left(\left(3h + 2\right) + 1\right)}^{2} - 9{\left(\left(3k - 4\right) - 2\right)}^{2} = 144$

$⇒ 144 ( h + 1{)}^{2} - 81 ( k - 2{)}^{2} = 144$

$⇒ 16\left({h}^{2} + 2h + 1\right) - 9\left({k}^{2} - 4k + 4\right) = 16$

$⇒ 16{h}^{2} - 9{k}^{2} + 32h + 36k - 36 = 0$

So, the locus of the centroid is

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