The correct option is D √133
Here, 2y−x−3 and 2x+y−1 are perpendicular to each other.
Therefore, the equation of the conic can be written as
4×5(2y−x−3√22+12)2−9×5(2x+y−1√22+12)2=80
⇒4(2y−x−3√5)2−9(2x+y−1√5)2=16
On putting 2y−x−3√5=X and 2x+y−1√5=Y, the given equation can be written as
4X2−9Y2=16
⇒X24−Y2(4/3)2=1
This is a Hyperbola.
Therefore, eccentricity is given by,
e=√1+16/94
=√133