The curve represented by the equation px+qy=1,where p,q∈R,p,q>0 is

We have px+qy=1or  (px+qy)2=1or  px+qy+2(pq)(xy)=1or  (px+qy−1)2=4(pq)(xy)or  p2x2−2(pq)(xy)+q2y2−2px−2qy+1=0On comparing this equation with the equationax2+2hxy+by2+2gx+2fy+c=0we 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=0Thus, we have Δ≠0 and h2=abHence, the given curve is a parabola.