# Tangent line to an implicitly defined cubic curve

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.

• I'm getting $\dfrac{dy}{dx}=\dfrac{12x-6xy-3y^2+10y-8}{3x^2+6xy+3y^2-10x-14y+8}$. – Taylor Nov 8 '16 at 12:54
• But would the solution set still not be infinite(a conic) when it should be finite? – Padraig Stapleton Nov 8 '16 at 13:16
• 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. – Taylor Nov 8 '16 at 13:35