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Given the curve defined implicitly below and a point $(1,1)$ on that curve I was asked to find the tangent line at that point. You then have to find all other tangent lines parallel to that one. $$ x^2 +3x^2y+3xy^2+y^3-7x^2-10xy-7y^2+8x+8y =0. $$

My attempt: I took the derivative with respect to $x$ of both sides and isolated $dy/dx$. I found this to be equal to $$ \frac{10y+14x-3x^2-6xy-8}{3y^2+3x^2-10x-14y+8}. $$ To find the slope at the given point, I simply substituted in $x = 1$ and $y = 1$ and found $\frac{dy}{dx}$ to be equal to $-0.7$. To find all other tangent lines I set the second equation equal to $-0.7$.

But here is my problem. That equation will give a solution set in the form of a conic, which implies there is an infinite amount of solutions, but a plot of the graph done online shows there cannot be. Many of the points given by the second equation are not on the original curve.

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  • $\begingroup$ I'm getting $\dfrac{dy}{dx}=\dfrac{12x-6xy-3y^2+10y-8}{3x^2+6xy+3y^2-10x-14y+8}$. $\endgroup$ – Taylor Nov 8 '16 at 12:54
  • $\begingroup$ But would the solution set still not be infinite(a conic) when it should be finite? $\endgroup$ – Padraig Stapleton Nov 8 '16 at 13:16
  • $\begingroup$ The solution set of $\dfrac{dy}{dx}=\text{whatever}$ will be infinite, but the extra condition that the solutions need to actually be on the given curve will make it finite, I think. $\endgroup$ – Taylor Nov 8 '16 at 13:35

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