Given conic is 12x2+16y2=1
e=1−1612=21
Foci (0,2) and (0,−2).
So, transverse axis of hyperbola=2b=4.
⇒b=2
Given e=23
And a2=b2(e2−1)
⇒a2=4(49−1)
⇒a2=5
∴The equation of hyperbola is 5x2−4y2=−1.
Now, Check options.
Clearly, the point (5,23) does not satisfy above equation.