Any point on the curve xny=an is,
P(at,1tn)
Differentiate w.r.t x, we get
nxn−1y+xndydx=0
⇒dydx=−nyx
⇒dydx=−natn+1
The equation of the tangent at P(at,1tn) is
y−1tn=−natn+1(x−at)
This meets the co-orinate axes at
A(at(n+1)n,0)B(0,(n+1)tn)
Area of ΔAOB=12(OA×OB)=12(at(n+1)n)((n+1)tn)=a(n+1)22nt1−n
For the area to be a constant,
1−n=0⇒n=1