Given, curves are c1≡xy=k and c2≡x=y2 On solving Eqs. (i) and (ii), we get the intersection point (k2∕3,k1∕3). On differentiating Eqs. (i) and (ii) w.r.t. x, we get ‌‌ for Eq. (i), ‌‌‌x‌
dy
dx
+y=0 ‌⇒‌‌(‌
dy
dx
)c1=‌
−y
x
. . . (iii) for Eq. (ii), ‌1=2y‌
dy
dx
‌⇒‌‌(‌
dy
dx
)c2=‌
1
2y
. . . (iv) ‌‌ At point ‌(k2∕3,k1∕3),(‌
dy
dx
)c1=−‌
k1∕3
k2∕3
. ‌‌ and ‌‌‌(‌
dy
dx
)c2=‌
1
2k1∕3
Since, the curves cut at right angle, then ‌(‌