The correct option is B y=3√27x+15√7
Given curves are 9x2−16y2=144 and x2+y2=9.
Let the equation of common tangent be y=mx+c.
Since, y=mx+c is a tangent to x2+y2=9,
∣∣∣0−0+c√1+m2∣∣∣=3⇒c2=9(1+m2) …(1)
Also, y=mx+c is a tangent to x216−y29=1
⇒c2=16m2−9 …(2)
(using condition of tangency)
From (1) and (2), we get
16m2−9=9+9m2⇒m2=187⇒m=±3√27
Also, c=±15√7
Hence, y=3√27x+15√7 is one of their common tangent.