Answer:
- tangent: y = 2x -2
- normal: y = -1/2x +3
Explanation:
Differentiating implicitly, you have ...
-y²·dx +(4-x)(2y)dy = 3x²·dx
So, the slope is ...
dy/dx = (3x² +y²)/(2y(4 -x))
At (x, y) = (2, 2), the slope of the curve is ...
dy/dx = (3·2² +2²)/(2·2(4 -2)) = 16/8 = 2
In point-slope form, the equation of the tangent line is then ...
y = m(x -h) +k
y = 2(x -2) +2
y = 2x -2 . . . . equation of tangent line
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The normal to the curve is perpendicular to the tangent at the same point. The slope of the perpendicular line is the negative reciprocal of the tangent's slope, so is -1/2.
y = (-1/2)(x -2) +2
y = -1/2x +3 . . . equation of normal line