The locus of the centroid of the triangle for | vedclass

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Question

Given hyperbola is

$16(x+1)^{2}-9(y-2)^{2}=164+16-36=144$

$\Rightarrow \frac{(x+1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1$

$	ext { Eccentricity, } e

Vedclass · Accepted Answer

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

Vedclass · Answer

Given hyperbola is

$16(x+1)^{2}-9(y-2)^{2}=164+16-36=144$

$\Rightarrow \frac{(x+1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1$

$	ext { Eccentricity, } e=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$

$\Rightarrow 	ext { foci are }(4,2) 	ext { and }(-6,2)$

Let the centroic be $(\mathrm{h}, \mathrm{k})$

$\, A(\alpha, \beta)$ be point on hyperbola

So $h=\frac{\alpha-6+4}{3}, k=\frac{\beta+2+2}{3}$

$\Rightarrow \alpha=3 h+2, \beta=3 k-4$

$(\alpha, \beta)$ lies on hyperbola so

$16(3 h+2+1)^{2}-9(3 k-4-2)^{2}=144$

$\Rightarrow 144(h+1)^{2}-81(k-2)^{2}=144$

$\Rightarrow 16\left(h^{2}+2 h+1ight)-9\left(k^{2}-4 k+4ight)=16$

$\Rightarrow 16 x^{2}-9 y^{2}+32 x+36 y-36=0$

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

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