We have, equarion of hyperbola ‌5x2−ky2=12 ‌‌
x2
12
5
−‌
y2
12
=1 Here, a2=‌
12
5
and b2=‌
12
K
Equation of tangent of hyperbola is 5x−2y−6=0 or y=‌
5
2
x−3 Here, m=‌
5
2
and a2m2−b2=9 ‌
12
5
×‌
25
4
−‌
12
K
=9⇒15−9=‌
12
K
K=2 Equation of hyperbola is ‌5x2−2y2=12 ‌‌
x2
12
5
−‌
y2
6
=1 Since, (√6,p) lie on hyperbola ‌⇒5(√6)2−2(p3)=12 ‌⇒‌‌30−2p2=12 ‌‌‌p2=9‌‌p=23 ‌∵‌‌p<0 ‌∴‌‌p=−3 Equation of normal of hyperbola of (√6,−3),3 ‌⇒‌‌‌