009A Sample Final 2, Problem 4

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

$3x^{2}+xy+y^{2}=5$  at the point  $(1,-2)$

Foundations:
The equation of the tangent line to  $f(x)$  at the point  $(a,b)$  is
$y=m(x-a)+b$  where  $m=f'(a).$

Solution:

Step 1:
We use implicit differentiation to find the derivative of the given curve.
Using the product and chain rule, we get
$6x+xy'+y+2yy'=5.$
We rearrange the terms and solve for  $y'.$
Therefore,
$xy'+2yy'=5-6x-y$
and
$y'={\frac {5-6x-y}{x+2y}}.$

Step 2:
Therefore, the slope of the tangent line at the point  $(1,-2)$  is
${\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {5-6(1)-(-2)}{1-4}}\\&&\\&=&\displaystyle {{\frac {1}{-3}}.}\end{array}}$
Hence, the equation of the tangent line to the curve at the point  $(1,-2)$  is
$f(x)=-{\frac {1}{3}}(x-1)-2.$

Final Answer:
$f(x)=-{\frac {1}{3}}(x-1)-2.$

Return to Sample Exam