Question

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

• A. $16x^2-9y^2+32x+36y-36=0$
• B. $9x^2-16y^2+36x+32y-144=0$
• C. $9x^2-16y^2+36x+32y-36=0$
• D. $16x^2-9y^2+32x+36y-144=0$

Answer 1

The correct option is A $16x^2-9y^2+32x+36y-36=0$

$H\equiv 16(x^2+2x+1)-9(y^2-4y+4)=144$
$\equiv \frac{(x+1{)}^2}{9}-\frac{(y-2{)}^2}{16}=1$
$e=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$
Foci $(\pm ae-1,2)=(\pm 5-1,2)$
Foci are $(4,2)$ and $(-6,2)$
Let a general point be $P(-1+3\sec \theta ,2+4\tan \theta )$
Let centroid be $(h,k)$
$h=\frac{4-6-1+3\sec \theta }{3},k=\frac{2+2+2+4\tan \theta }{3}$
$\Rightarrow h=-1+\sec \theta$ and
$3k=6+4\tan \theta$
$\Rightarrow \sec \theta =(h+1)$ and $\frac{3}{4}(k-2)=\tan \theta$
Now ${\sec }^2\theta -{\tan }^2\theta =1$
$\Rightarrow 16(h+1{)}^2-9(k-2{)}^2=16$
$\Rightarrow 16x^2-9y^2+32x+36y-36=0$