First, we find the equation of tangent to the hyperbola x2−16y2−2x−64y−72=0 . . . (i) at the point (−4,−3), which is given by ‌−4x+48y−(x−4)−32(y−3)−72=0 ‌⇒−4x+48y−x+4−32y+96−72=0 ‌⇒−5x+16y+28=0 . . . (ii) Required normal is the normal to the tangent (ii), which is given by 16x+5y+c=0 But it passes through (−4,−3). ‌∴‌−64−15+c‌=0 . . . (iii) ⇒c‌=79 Putting this value in Eq. (iii), we get 16x+5y+79=0 Which is the required normal.