Question

I need help with this question:

Find all the x-coordinates of the points on the curve x^2y^2+xy=2 where the slope of the tangent line is −1.

Answer 1

I've used implicit differentiation to get

\[2xy^2+x^22yy'+y+xy=0\]

I then substituted y' with -1 and got

\[2xy^2-x^22y+y-x\]

Answer 2

I don't know what to do next...

Answer 3

so you have
\[2xy^2+x^22yy'+y+xy'=0 \\ \text{ okay and you replaced } y' \text{ with } -1 \\ 2xy^2-2x^2y+y-x=0 \\ 2xy(y-x)+1(y-x)=0 \\ (y-x)(2xy+1)=0 \\ \text{ so we have } y=x \text{ or } xy=\frac{-1}{2} \\ \text{ so try plugging both of these possibilities back in the original equation }\]

Answer 4

So wherever I have a y I replace with x, and wherever I have xy I replace with -1/2 ?

Answer 5

yeah we will see one equation is never true

Answer 6

and you will see one equation is true for 2 real values

Answer 7

2 real x values

Answer 8

for example
with the xy=-1/2 sub
\[(xy)^2=(\frac{-1}{2})^2 \implies x^2y^2=\frac{1}{4} \\ \text{ but } \frac{1}{4}+\frac{-1}{2} \text{ is never } 2 \]

Answer 9

you can solve the one with y=x pretty easily

Answer 10

Hmmm... but replacing y with x I get ...

x^4 + x^2 = 2

Answer 11

right that is a quadratic in terms of x^2

Answer 12

do you know how to solve u^2+u=2?

Answer 13

Yes! Oh so now I see it

Answer 14

I replaced x^2 with u
so we can solve for u
then replace u with x^2
and then solve for x

Answer 15

Ah thank you so much @freckles !!! I got -1, and 1 and it's correct!

Answer 16

great! :)

Answer 17

I appreciate the help! ^.^

Answer 18

np