If the pair of lines 3x2−5xy+py2=0 and 6x2−xy−5y2=0 have one line in common, then p=
6x2−xy−5y2=0
Substitute t=xy
6t2−t−5=0
⇒t=1±√1+12012
⇒t=1±1112
⇒t=1,−56
y=x or y=−65x
∴3x2−5xy+py2=0
t=xy, 3t2−5t+p=0
If, t=1,p=2
And If, t=−56,p=5×(−56)−3×2536
=−25×636−7536
=−22536
p=−254
So, p=2 or −254 are two values.