Question

4x^2 + (y-2)^2 find dy/dx
find the equation of the line tangent to the curve at point (0,6)

Answer 1

dy/dx = 8x + 2(y-2)* dy/dx
2(y-2)* dy/dx - dy/dx = -8x
dy/dx = -8x / [ 2(y-2) - 1]
= -8x ( 2y -5)

Answer 2

so that first part is equal to 16 so do i just do -8x(2y-5)=16 then solve for y?

Answer 3

the 4x^2 + (y-2)^2 = 16 ?

Answer 4

what was the original question?