2x + 2y + 2xy' - 2yy' + 1 = 0
(2x+2y+1) + y'(2x-2y) = 0
y' = (2x+2y+1)/(2y-2x)
y' at (3,5) = (6+10+1)/(10-6) = 17/4
so, you want the line through (3,5) with slope 17/4:
(y-5) = 17/4 (x-3)
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
x2 + 2xy − y2 + x = 17, (3, 5)
(hyperbola)
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