The correct option is C a hyperbola
12x2−23xy+10y2−25x+26y−14=0
On comparing with ax2+2hxy+by2+2gx+2fy+c=0, we have
a=12, h=−232,b=10,g=−252,f=13,c=−14
⇒h2=5294, ab=120
⇒ h2>ab ⋯(i)
Now, △=abc+2fgh−af2−bg2−ch2
=12(10)(−14)+2(13)(−252)(−232)−12(13)2−10(−252)2−(−14)(−232)2
=6372≠0 ⋯(ii)
From (i) and (ii), the above equation represents a hyperbola.