# Tangent line off a curve

I'm having trouble figuring out how to find the equation to a tangent line off a curve. I know you use implicit differentiation of the function, but lets say the derivative found is still a difficult function, such as $$-2x + y^2/-2xy + 1$$ Now, the formula for the tangent line is y = mx + c and lets say it passes through the point (-3, -5). How would you go about figuring this out?

• Is (-3,-5) the point of tangency? – S. Dolan Sep 7 at 7:45
• Do you mean $\dfrac{-2x+y^2}{-2xy+1}$? – José Carlos Santos Sep 7 at 7:52
• Yes I mean that – DuncanK3 Sep 8 at 4:03

Firstly you find $$\frac {dy}{dx}$$. For this you only need to use implicit differentiation if you do not have an explicit formula for $$y$$. Then substitute the coordinates of the point of tangency into your expression for $$\frac {dy}{dx}$$.
Example $$\frac {-2x+y^2}{-2xy+1}=\frac {-31}{29}$$ at $$(-3,-5)$$.
This gradient is the $$"m"$$ in $$y=mx+c$$.
Example $$y=\frac {-31}{29}x+c$$. You then use the point $$(-3,-5)$$ to find $$"c"$$.
• That's right. At a point of tangency the gradient of the curve is the same as the gradient $m$ of the line. – S. Dolan Sep 8 at 7:41