#xy = a(y+√(y²-x²))# then show that #x³dy/dx=y²(y+√(y²-x²))# ?

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Answer 1 · 2016-10-01T23:25:27

#dy/dx = (y^2)/(x^3)(y+sqrt(y^2-y^2))#

Making #f(x,y) = (xy)/(y+sqrt(y^2-x^2))-a=0#

we have

#df = f_x dx + f_y dy = 0#

with

#f_x = y^2/(-x^2 + y (y + sqrt[-x^2 + y^2]))#
#f_y = (x (y - sqrt[-x^2 + y^2]))/(x^2 - y (y + sqrt[-x^2 + y^2]))#

so

#dy/dx = -(f_x)/(f_y) = (y^2)/(x^3)(y+sqrt(y^2-y^2))#

#x*y = a(y+√(y²-x²))# then show that #x³*dy/dx=y²(y+√(y²-x²))# ?