1No replies!!!!!!!!!!!
6 Answers
62this i think is from A Das Gupta..
-x/√1-x2 - y/√1-y2 (dy/dx) = a(1-dy/dx)
-x/√1-x2 - y/√1-y2 (dy/dx) = (√1-x2 + √1-y2)(1-dy/dx) / (x-y)
(√1-x2 + √1-y2) / (x-y) + x/√1-x2 = - y/√1-y2 (dy/dx) + (√1-x2 + √1-y2)(dy/dx) / (x-y)
(1-x2 + √1-y2√1-x2 + (x-y)x )/√1-x2 = (- y(x-y) + (√1-x2√1-y2 + 1-y2)(dy/dx) /√1-y2
(1-x2 + √1-y2√1-x2 + (x-y)x )/√1-x2 = (- y(x-y) + (√1-x2√1-y2 + 1-y2)(dy/dx) /√1-y2
(1-yx + √1-y2√1-x2 )/√1-x2 = ( -yx + 1 + (√1-x2√1-y2 ) dy/dx) /√1-y2
Thus,
dy/dx=√1-y2/√1-x2
1bhaiyya my sir gave as homework which when solved not getting..........
62I thought that first.. then i realised that this was possible like this too :)
1let x=sinθ and x=sinφ
θ=sin-1x and φ=sin-1y
substituing..........
u'll get
√1-sin2θ+√1-sin2φ=a(sinθ-sinφ)
so
√cos2θ + √cos2φ=a(sinθ-sinφ)
a=(cosθ+cosφ)/(sinθ-sinφ)
applying properties.........
a= (2 cos (θ+φ)/2 cos(θ-φ)/2) / (2 cos (θ+φ)/2 sin(θ-φ)/2)
= cos (θ+φ)/2 sin(θ-φ)/2
= cot(θ+φ)/2
2cot-1a=θ+φ
sinx+siny=2cot-1a
Differentiating..........
1/√1-x2 -1/√1-y2 dy/dx= 0
1/√1-x2 =1/√1-y2 dy/dx
Thus.............
dy/dx=1/√1-x2/1/√1-y2
dy/dx=√1-y2/√1-x2