If the curve y^2 = x and xy = k are orthogonal then prove that 8k^2 = 1.
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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
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
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