# Thread: Finding a tangent line

1. ## Finding a tangent line

What would be the equations of the tangent lines to the graph at the following info:

$\displaystyle 4xy+x^2=5$ at the point $\displaystyle (1,1)$ and $\displaystyle (5,-1)$

If you can show me a step by step solution that would be easiest for me to follow.

2. Originally Posted by AMaccy
What would be the equations of the tangent lines to the graph at the following info:

$\displaystyle 4xy+x^2=5$ at the point $\displaystyle (1,1)$ and $\displaystyle (5,-1)$

If you can show me a step by step solution that would be easiest for me to follow.
Use "implicit differentiation" to find y': $\displaystyle 4y+ 4xy'+ 2x= 0$. $\displaystyle 4xy'= -4y- 2x$ and $\displaystyle y'= \frac{-4y-2x}{4x}$

At (1, 1), $\displaystyle y'= \frac{-4(1)- 2(1)}{4(1)}= \frac{-6}{4}= -\frac{3}{2}$. The tangent line is $\displaystyle y= -\frac{3}{2}(x- 1)+ 1$.

At (5, -1), $\displaystyle y'= \frac{-4(-1)- 2(5)}{4(5)}= \frac{4- 10}{20}= \frac{-6}{20}= -\frac{3}{10}$. The tangent line is $\displaystyle y= -\frac{3}{10}(x- 5)- 1$.