Given, xnym=am+n,m,n>0
Taking logarithm on both sides, we get
log (xnym)=logam+n ⇒logxn+logym=(m+n) log a ⇒nlogx+mlogy=(m+n)loga...(i)
On differentiating Eq. (i) w.r.t. 'x', we get xn+ym=0 ⇒ymdxdy=−xn ⇒dxdy=−(mn)(xy) ∴ Length of subtangent =dy/dxy =−(mn)(xy)y =n−mx ∴ Length of sub tangent at (x1,y1)=nm∣x1∣