Question

find the equation of tangent to the curve (x\a)^n + (y\b)^n = 2 at the point (a, b)

Answer 1

assuming a and b are constants, we have

(1/a^n) x^n + (1/b^n) y^n = 2
(n/a^n) x^(n-1) + (n/b^n) y^(n-1) dy/dx = 0

so for (a,b)
(n/a^n)(a^(n-1) + (n/b^n)(b^(n-1) dy/dx = 0
(n/a) + (n/b) dy/dx = 0
dy/dx = -(n/a) (b/n) = -b/a

equation of tangent at (a,b)
y - b = (-b/a)(x-a)

toy with this any way you want to