SOLUTION: Find the equation of the tangent and normal to x^2 + 3xy + y^2 = 5 at P(1,1).

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Question 1082777: Find the equation of the tangent and normal to x^2 + 3xy + y^2 = 5 at P(1,1).
Answer by Alan3354(69443) About Me  (Show Source):
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Find the equation of the tangent and normal to x^2 + 3xy + y^2 = 5 at P(1,1).
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Find the 1st derivative
x^2 + 3xy + y^2 = 5
Differentiate implicitly.
2xdx + 3ydx + 3xdy + 2ydy = 0
(2x+3)dx + (2y+3)dy = 0
dy/dx = -(2x+3)/(2y+3) = m, the slope
At (1,1) m = -1
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At (1,1), y-1 = -(x-1)
y = -x + 2 is tangent.
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y = x is normal