The locus of the centroid of the triangle for | vedclass

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(B) The given hyperbola is $16(x+1)^{2}-9(y-2)^{2}=164+16-36=144$.This simplifies to $\frac{(x+1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1$.The center of the hy

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

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(B) The given hyperbola is $16(x+1)^{2}-9(y-2)^{2}=164+16-36=144$.This simplifies to $\frac{(x+1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1$.The center of the hyperbola is $(-1, 2)$.Here $a^{2}=9$ and $b^{2}=16$,so $e=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$.The foci are $(h \pm ae, k) = (-1 \pm 3 	imes \frac{5}{3}, 2) = (-1 \pm 5, 2)$,which are $(4, 2)$ and $(-6, 2)$.Let $P(\alpha, \beta)$ be a point on the hyperbola.Let $G(x, y)$ be the centroid of the triangle formed by $P(\alpha, \beta)$,$(4, 2)$,and $(-6, 2)$.Then $x=\frac{\alpha+4-6}{3} = \frac{\alpha-2}{3} \Rightarrow \alpha=3x+2$.And $y=\frac{\beta+2+2}{3} = \frac{\beta+4}{3} \Rightarrow \beta=3y-4$.Since $P(\alpha, \beta)$ lies on the hyperbola $16(x+1)^{2}-9(y-2)^{2}=144$,we substitute $\alpha$ and $\beta$:$16(3x+2+1)^{2}-9(3y-4-2)^{2}=144$$16(3x+3)^{2}-9(3y-6)^{2}=144$$16 	imes 9(x+1)^{2}-9 	imes 9(y-2)^{2}=144$$144(x+1)^{2}-81(y-2)^{2}=144$Divide by $9$:$16(x+1)^{2}-9(y-2)^{2}=16$$16(x^{2}+2x+1)-9(y^{2}-4y+4)=16$$16x^{2}+32x+16-9y^{2}+36y-36=16$$16x^{2}-9y^{2}+32x+36y-36=0$.

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

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