The correct option is C a hyperbola if λ<4
Comparing the given equation with standard conic equation
ax2+2hxy+by2+2gx+2fy+c=0, we get
a=λ=2g,h=2,b=1,f=3/2,c=2
Now, Δ=abc+2fgh−af2−bg2−ch2
⇒Δ=14(−λ2+11λ−32)
Hence Δ≠0, ∀ λ∈R
Now, for conic to be parabola,
22=1⋅λ⇒λ=4
For conic to be ellipse, λ>4
and for conic to be hyperbola, λ<4