The equation of the tangent line to the curve at the given point. x2 + y2 = (2x2+2y2 + 7x)2, (0, 1) is y =

Answer 1

I interpret your equation as

x² + y² = (2x² + 2y² + 7x)²

The tangent will be of the form

y = px + q, where

p = dy/dx (also denoted y') and

q will be the result of constraining the line to go though the point (0, 1).

To find dy/dx by implicit differentiation, we differentiate the equation by dx,

while treating y as a function of x:

2x + 2yy' = 2(2x² + 2y² + 7x)(4x + 4yy' + 7)

At the point x = 0, y = 1:

2y' = 2(2)(4y' + 7)

2y' = 16y' + 28

y' = -2

So the tangent is of the form

y = -2x + q

At the point x = 0, y = 1:

1 = q

So the tangent is

y = -2x + 1