The equation 16x2−3y2−32x+12y−44=0 represents a hyperbola
the length of whose transverse axis is 43
the length of whose conjugate axis is 4
whose center is (−1,2)
whose eccentricity is 19/3
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).