Answer :
Answer:
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
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