Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
2x2 + xy + 2y2 = 5, (1, 1)
(ellipse)

Question

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
2x2 + xy + 2y2 = 5,  (1, 1)
  (ellipse)

anonymous · Answer

$$2x^2+xy+2y^2=5$$
Implicit differentiation yields
$$4x+y+xy'+4y~y'=0$$
Solve for $y'$.

anonymous · Answer

ok let me try it quick

anonymous · Answer

i got :|dw:1370917766115:dw|

anonymous · Answer

Not quite:
$$4x+y+xy'+4y~y'=0\
y'~(x+4y)=-(4x+y)\
y'=-\frac{4x+y}{x+4y}$$

anonymous · Answer

my apologies- i see where i made my mistake...would this be my final answer

anonymous · Answer

Nope. $y'$ gives the slope of the tangent line at any given point. Given the point $(1,1)$, what would $y'$ be?

anonymous · Answer

so it would be a - 3/5 slope?