If -x^2y-4+4y-2x=0−x
2
y−4+4y−2x=0 then find the equations of all tangent lines to the curve when y=2.y=2.

y = 2x

Step-by-step explanation:

-x² y − 4 + 4y − 2x = 0

Take the derivative of both sides with respect to x.

-x² dy/dx + (-2x) y + 4 dy/dx − 2 = 0

Solve for dy/dx.

-2xy + (4 − x²) dy/dx − 2 = 0

(4 − x²) dy/dx = 2xy + 2

dy/dx = (2xy + 2) / (4 − x²)

Now use the original equation to find x when y = 2.

-x² (2) − 4 + 4(2) − 2x = 0

-2x² − 2x + 4 = 0

x² + x − 2 = 0

(x − 1) (x + 2) = 0

x = -2 or 1

Plug back in to check for extraneous solutions.

-(-2)² y − 4 + 4y − 2(-2) = 0

-4y − 4 + 4y + 4 = 0

0 = 0

y = all numbers

-(1)² y − 4 + 4y − 2(1) = 0

-y − 4 + 4y − 2 = 0

3y − 6 = 0

y = 2

x = -2 is an extraneous solution.  Only x = 1 is a solution.

The point is (1, 2).  Plug this into the first derivative to find the slope of the tangent line.

dy/dx = (2(1)(2) + 2) / (4 − (1)²)

dy/dx = (4 + 2) / (4 − 1)

dy/dx = 2

Now write the equation using point-slope form.

y − 2 = 2 (x − 1)

If desired, convert to slope-intercept form.

y = 2x

Graph: desmos.com/calculator/tsmjxjfqbo