# 009A Sample Final 2, Problem 4

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

${\displaystyle 3x^{2}+xy+y^{2}=5}$  at the point  ${\displaystyle (1,-2)}$
Foundations:
The equation of the tangent line to  ${\displaystyle f(x)}$  at the point  ${\displaystyle (a,b)}$  is
${\displaystyle y=m(x-a)+b}$  where  ${\displaystyle m=f'(a).}$

Solution:

Step 1:
We use implicit differentiation to find the derivative of the given curve.
Using the product and chain rule, we get
${\displaystyle 6x+xy'+y+2yy'=0.}$
We rearrange the terms and solve for  ${\displaystyle y'.}$
Therefore,
${\displaystyle xy'+2yy'=-6x-y}$
and
${\displaystyle y'={\frac {-6x-y}{x+2y}}.}$
Step 2:
Therefore, the slope of the tangent line at the point  ${\displaystyle (1,-2)}$  is
${\displaystyle {\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {-6(1)-(-2)}{1-4}}\\&&\\&=&\displaystyle {{\frac {4}{3}}.}\end{array}}}$
Hence, the equation of the tangent line to the curve at the point  ${\displaystyle (1,-2)}$  is
${\displaystyle y={\frac {4}{3}}(x-1)-2.}$

${\displaystyle y={\frac {4}{3}}(x-1)-2.}$