# Thread: Implicit Diff and an Ellipse

1. ## Implicit Diff and an Ellipse

Consider the graph of x^2+xy+y^2=1. Use implicit differentiation to find all points on the graph where the tangent line is parallel to the line y=-x.

Where do I go after I implicitly differentiate this thing?

2x+y'+2y*y'=0
y'+2y*y'=-2x
y'(1+2y)=-2x
y'=-(2x)/(1+2y)

Did I do that right?

2. Originally Posted by ebonyscythe
Consider the graph of x^2+xy+y^2=1. Use implicit differentiation to find all points on the graph where the tangent line is parallel to the line y=-x.

Where do I go after I implicitly differentiate this thing?

2x+y'+2y*y'=0
y'+2y*y'=-2x
y'(1+2y)=-2x
y'=-(2x)/(1+2y)

Did I do that right?
you did not do this correctly. note that you need the product rule to differentiate the xy

when you get the correct derivative, solve for y' and set it equal to -1 (the slope of y = -x)

3. Hello, ebonyscythe!

Your differentiation is off . . .

Consider the graph of $\displaystyle x^2+xy+y^2\:=\:1$
Use implicit differentiation to find all points on the graph
where the tangent line is parallel to the line $\displaystyle y =-x$

$\displaystyle \text{We have: }\;x^2 + \underbrace{xy}_{\text{product}} + y^2\:=\:1\;\;{\color{blue}[1]}$
. . $\displaystyle \text{then: }\;2x + \overbrace{xy' + y} + 2yy' \:=\:0$

Then we have: .$\displaystyle xy' + 2yy' \:=\:-2x - y$

Factor: .$\displaystyle (x + 2y)y' \:=\:-(2x + y)$

. . Hence: .$\displaystyle y' \:=\:-\frac{2x + y}{x + 2y}$

$\displaystyle \text{The slope of }\, y \,= \,\text{-}x\:\text{ is }-1.$

Set: .$\displaystyle -\frac{2x + y}{x + 2y} \:=\:-1\quad\Rightarrow\quad y \,=\,x$

Substitute into [1]: .$\displaystyle x^2 + x\!\cdot\!x + x^2\:=\:1\quad\Rightarrow\quad3x^2\:=\:1\quad\Righ tarrow\quad x \:=\:\pm\frac{\sqrt{3}}{3}$
. . Then: .$\displaystyle y \:=\:\pm\frac{\sqrt{3}}{3}$

Answers: .$\displaystyle \left(\frac{\sqrt{3}}{3},\:\frac{\sqrt{3}}{3}\righ t),\;\left(\text{-}\,\frac{\sqrt{3}}{3},\:\text{-}\,\frac{\sqrt{3}}{3}\right)$