Question

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

Answer 1

`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))`