Question

Help please!!!! Rotated Ellipse with the equation x^(2)+xy+y^(2)=1. Find all the points where the tangent line is horizontal. Then find all points where the tangent line is parallel to the line y=-x.

Answer 1

so first question
find for what x values we have y'=0
second question
find for what x values we have the y'=-1

Answer 2

or all (x,y) such that y'=0 (question 1)
and find all (x,y) such that y'=-1 (question 2)

Answer 3

have you found y' yet?

Answer 4

Yeah, y'= (-2x-y)/(x+2y)

Answer 5

\[\text{ so } y'=0 \text{ when } -2x-y=0 \\ \text{ and } y'=-1 \text{ when } \frac{-2x-y}{x+2y}=-1\]

Answer 6

It is easy to solve that first equation for y
it is a little more difficult to solve that bottom equation for y but not impossible

Answer 7

also you should probably exclude when x+2y can be zero from both answers

Answer 8

when you get y=-2x what do you do to get the points that make it horizontal? And I'm not sure how to go about solving the second one. :/

Answer 9

hey,
you found y = -2x
but we want the points on the ellipse so we plug y=-2x to the ellipse equation:
x^2 + x * (-2x) + (-2x)^2 = 1
x^2 -2x^2 +4x^2 = 1
x^2 = 1/3
so you have two values of x..

Answer 10

So +/- sqrt(1/3)?

Answer 11

yes x values of the points so you can find now the y values as well

Answer 12

what would you do for that?

Answer 13

you can use either the ellipse equation or y=-2x

Answer 14

Oh okay

Answer 15

do you understand why we have to plug y=-2x into the ellipse equation ?

Answer 16

To find the points that make the tangent line horizontal.

Answer 17

when we took the derivative of the ellipse we got y=-2x.
now since this derivative may be the same for different ellipses (for example with other values then 1 in the right side) we couldnt find the points coordinates.
we had to put this result in the specific ellipse equation

Answer 18

i hope it make sense

Answer 19

To find the points where the tangent line is parallel to the line y=-x what would you do??

Answer 20

use the derivative and find a relation between y and x
then again plug it into the ellipse equation

Answer 21

the slope is -1 would you set the derivative equal to that?

Answer 22

correct

Answer 23

is that x=-y?

Answer 24

i think y=x

Answer 25

(-2x-y)/(x+2y) = -1
-2x-y=-x-2y
2y-y=2x-x
y=x

Answer 26

then just plug that into the ellipse equation?

Answer 27

thanks for your help! :)

Answer 28

yes then you will find values of x\y and then you will find the points

Answer 29

sure no problem

Answer 30

is it x=1/3 again?

Answer 31

well, the sqrt1/3

Answer 32

so then the points would have the same x and y values?

Answer 33

yes i think so, but you will get different y values

Answer 34

wouldn't they be the same because y=x?

Answer 35

sorry i meant different from the first case y'=0

Answer 36

for y'=-1 the x and y are the same indeed

Answer 37

(sqrt1/3,sqrt1/3)(-sqrt1/3,-sqrt1/3)?

Answer 38

yes :) the points for y'=-1

Answer 39

Okay! Thanks again.

Answer 40

yw