Comparing 3ax2−16xy−(a2−10)y2=0 with Ax2+2Hxy+By2=0 A=3a,H=−8‌ and ‌B=−(a2−10) If lines are perpendicular, then A+B‌‌=0 3a−a2+10‌‌=0 ‌‌ or ‌‌‌a2−3a−10‌‌=0 ‌‌ or ‌‌‌(a+2)(a−5)‌‌=0 ⇒‌‌‌a‌‌=−2‌ or ‌a=5 ∴ Lines are perpendicular when a=−2