JEE Main 2021 — Hyperbola Question with Solution

Given hyperbola is

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

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

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

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

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

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

$\therefore$ Eccentricity, $e=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$

We know that, focii are $\left(h+ae, k\right)$ and $\left(h-ae, k\right)$

Hence, focii are $(4,2)$ and $(-6,2)$

Let the centroid be $(h,k)$ & $A(\alpha ,\beta )$ be point on hyperbola

So $h=\frac{\alpha -6+4}{3}, k=\frac{\beta +2+2}{3}$

$\Rightarrow \alpha =3h+2, \beta =3k-4$

Put the values of $\alpha ,\beta$ in the equation$\left(1\right)$,

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

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

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

$\Rightarrow 16h^2-9k^2+32h+36k-36=0$

So, the locus of the centroid is 

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