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# Find the point besides the origin where the graph of xy=x+y^2 has a vertical tangent

Starting with xy = x + y2, implicit differentiation gives

y + xy' = 1 + 2yy'

We need to solve for y':

xy' - 2yy' = 1 - y

(x - 2y)y' = 1 - y

y' = (1 - y) / (x - 2y)

The slope is vertical if y' is undefined, which happens if x - 2y = 0 so that x = 2y.

Substituting this into the original equation gives 2y2 = 2y + y2. We then solve for y

y2 - 2y = 0

y(y - 2) = 0

y = 0 or y = 2.

Thus the possible points are (0,0) and (4,2)

We must check that in the expression for y', the numerator is not 0 for each point. Indeed that is the case here. Thus both points work.

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