# Implicit Diff and an Ellipse

• Oct 15th 2007, 07:08 PM
ebonyscythe
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?
• Oct 15th 2007, 07:32 PM
Jhevon
Quote:

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)
• Oct 15th 2007, 07:46 PM
Soroban
Hello, ebonyscythe!

Your differentiation is off . . .

Quote:

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 $y =-x$

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

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

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

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

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

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

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

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