Answers

2014-05-21T14:28:45+05:30
Y² = x -------------(1) &
xy = k --------------(2) are orthogonal to each other hence tangent on are orthogonal
differentiating both equation 
2ydy/dx = 1
dy/dx = 1/2y     let m1
xdy/dx + y = 0
dy/dx = -y/x     let m2
for orthogonal 
m1*m2 = -1
(-y/x)*(1/2y) = -1
x = 1/2
put x = 1/2 in (2) equation
y = 2k
put y = 2k & x = 1/2 in equation (1)
4k² = 1/2
8k² = 1  prove
1 2 1