Question

The equation of the line tangent to (x^2+y^2)^4=16x^2y^2 at (x,y)=(−1,1) is ay = bx + c, where a, b and c are positive integers. What is the value of a+b+c?

Answer 1

(x^2+y^2)^4=16x^2y^2

(x^2+y^2)^3 (x+yy') = 4xy^2 + 4x^2yy'
x(x^2+y^2)^3 + y(x^2+y^2)^3 y' = 4xy^2 + 4x^2yy'
(y(x^2+y^2)^3 - 4x^2y)y' = 4xy^2 - x(x^2+y^2)^3
y' = (4xy^2 - x(x^2+y^2)^3)/(y(x^2+y^2)^3 - 4x^2y)

at (1,-1) y' = (4-8)/(-8+4) = 1

So, now you have a point and a slope, so

y+1 = 1(x-1)
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