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if x^2 + y^2 = xy + 3, find dy/dx where x = 1, and hence find the equations of the two tangents to the curve where x=1.
If \(\displaystyle x^2\,+\,y^2 \:= \:xy\,+\,3\), (a) find \(\displaystyle \frac{dy}{dx}\) where \(\displaystyle x\,=\,1\).
(b) Find the equations of the two tangents to the curve where \(\displaystyle x\,=\,1\).