The equation 16x2−3y2−32x+12y−44=0 represents a hyperbola

Let the equation of any normal be y=−tx+2t+t3.  Since it passes through the point (15,12), we have  12=−15t+2t+t3 or t3−13t−12=0 One root is −1. Then, (t+1)t2+t−12=0 or  t=1,3,4 Therefore, the co-normal points are (1,−2),(9,−6), and (16,8).  Therefore, the centroid is (26/3,0).