Question

prove tht the curve s x=y2 & xy=k cut at right angle if 8k2=1

Answer 1

both curves cut right angle when ,
products of slope of both lines =-1
hence,
x=y^2
differentiate
dx=2ydy
dy/dx=1/2y
let they cut at (a,b)
slope of curve x=y^2 is 1/2b

now ,
again we find slope of other curve
xy=k
ydx+xdy=0
dy/dx= -y/x
hence slope at (a,b) is -b/a

now both curve cut at (a,b)
so,
b^2=k/b
b^3=k hence a =k^2/3
now
product of slopes =1/2b.(-b/a)=-1
1/2a=1
1=2a
put a =k^2/3

hence. 1=2k^2/3
take cube both side
1=8k^2.
hence proved