How to solve this? please help me

Step-by-step 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

__

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

c is the final answer! : )

answer:      

step-by-step explanation: