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अतिपरवलय, $16 x ^{2}-9 y ^{2}+32 x +36 y -164=0$ पर किसी बिंदु $P$ तथा इसकी नाभियों से बने त्रिभुज के केन्द्रक का बिन्दुपथ है
$9 x^{2}-16 y^{2}+36 x+32 y-36=0$
$16 x^{2}-9 y^{2}+32 x+36 y-36=0$
$16 x^{2}-9 y^{2}+32 x+36 y-144=0$
$9 x^{2}-16 y^{2}+36 x+32 y-144=0$
Solution
Given hyperbola is
$16(x+1)^{2}-9(y-2)^{2}=164+16-36=144$
$\Rightarrow \frac{(x+1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1$
$\text { Eccentricity, } e=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$
$\Rightarrow \text { foci are }(4,2) \text { and }(-6,2)$
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
$16(3 h+2+1)^{2}-9(3 k-4-2)^{2}=144$
$\Rightarrow 144(h+1)^{2}-81(k-2)^{2}=144$
$\Rightarrow 16\left(h^{2}+2 h+1\right)-9\left(k^{2}-4 k+4\right)=16$
$\Rightarrow 16 x^{2}-9 y^{2}+32 x+36 y-36=0$
