The point on the curve with and positive –coordinate is

MATHEMATICS · Single correct · AP · Calculus AB · Differentiation: Composite, Implicit, and Inverse Functions · Differentiation: Composite, Implicit, and Inverse Functions

Problem

The point on the curve x2y+y2=9x^{2}y+y^{2}=9 with x=1x=1 and positive yy–coordinate is (1,−1+372)\Bigl(1,\tfrac{-1+\sqrt{37}}{2}\Bigr). What is the slope of the tangent line to the curve at that point? (a) 37−137\dfrac{\sqrt{37}-1}{\sqrt{37}} (b) −1-1 (c) 137−1\dfrac{1}{\sqrt{37}}-1 (d) 1−37237\dfrac{1-\sqrt{37}}{2\sqrt{37}}

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