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

1 Answer
Oct 1, 2016

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

Explanation:

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))