Question

Consider the equation: x^2+xy+y^2=1.
Find dy/dx
Find all points where the tangent line is horizontal.

Answer 1

I actually found the dy/dx: (-2x-y)/(x+2y)
I cannot figure out how to find the horizontal tangents by setting to zero. I'm having a hard time solving.

Answer 2

checking one sec

Answer 3

ok your y' looks great

Answer 4

horizontal lines have slope 0

Answer 5

so we need to find when y'=0

Answer 6

\[\frac{-2x-y}{x+2y}=0\]


by the way y' does not exist when x+2y=0

so find when -2x-y=0 and find when x+2y=0
so y'=0
when -2x-y=0 but exclude any numbers from the equation x+2y=0

Answer 7

if you are there
can you tell me what you get when you solve -2x-y=0 for y?

Answer 8

sorry, I was working on some other problems. ok, so I have y= -2x

Answer 9

right and solving the equation in which gives us y' does not exist
is x+2y=0
so y=-1/2 x
there

y=-1/2x and y=-2x
share a common point (0,0)
right?

Answer 10

so any point of the form (x,-2x)
where x is not zero is where you have y' is 0

Answer 11

does that make sense to you?

Answer 12

i must leave ya now
but good luck

Answer 13

oh i can wait a sec if you have something else to say

Answer 14

Thank You! I think I am understanding it now. Thanks!

Answer 15

I just found when the top is zero
and excluded any numbers in that set if they were also a number that made the bottom zero

Answer 16

number I meant point

Answer 17

anyways have fun! :)

Answer 18

and actually you might want to see if that is point
or see for which points (x,-2x) where x doesn't equal 0
satisfies
x^2+xy+y^2=1
x^2+x(-2x)+(-2x)^2=1
x^2-2x^2+4x^2=1
solve for x