The correct option is B a parabola
We have, √px+√qy=1
⇒(√px+√qy)2=1
⇒px+qy+2√(pq)(xy)=1
⇒(px+qy−1)2=4(pq)(xy)
⇒p2x2−2(pq)(xy)+q2y2−2px−2qy+1=0
On comparing this equation with the equation
ax2+2hxy+by2+2gx+2fy+c=0, we get
a=p2,b=q2,c=1,g=−p,
f=−q and h=−pq
Δ=abc+2fgh−af2−bg2−ch2
p2q2−2p2q2−p2q2−p2q2−p2q2
=−4p2q2≠0
and, h2−ab=p2q2−p2q2=0
Thus, we have Δ≠0 and h2=ab
Hence, the given curve is parabola.