A curve has equation:
x^3 - 2xy - 4x + y^3 - 51 = 0
find an equation of the normal to the curve at the point (4,3), giving your answer in the form: ax + by + c = 0, where a,b and c are integers.
x^3 - 2xy - 4x + y^3 - 51 = 0
find an equation of the normal to the curve at the point (4,3), giving your answer in the form: ax + by + c = 0, where a,b and c are integers.
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Differentiating implicitly:
3x^2 - 2xy' - 2y - 4 + 3y^2y' = 0
y' = (3x^2 - 2y - 4)/(2x - 3y^2)
Slope of tangent line at (4, 3) = (38)/(-19) = -2
Slope of normal line = 1/2
Eqn of normal line:
y - 3 = 1/2*(x - 4)
y = x/2 - 1
y - x/2 + 1 = 0
2y - x + 2 = 0
3x^2 - 2xy' - 2y - 4 + 3y^2y' = 0
y' = (3x^2 - 2y - 4)/(2x - 3y^2)
Slope of tangent line at (4, 3) = (38)/(-19) = -2
Slope of normal line = 1/2
Eqn of normal line:
y - 3 = 1/2*(x - 4)
y = x/2 - 1
y - x/2 + 1 = 0
2y - x + 2 = 0