Unable to find the intersection between a circle and ellipse

[lotur512]

Homework Statement

Find the intersection/system of equation between a parabola and circle

Relevant Equations

x^2+xy+y^2=18; x^2+y^2=12

Given:
x^2+xy+y^2=18
x^2+y^2=12

Attempt:
(x^2+y^2)+xy=18
12+xy=18
xy=6

y^2=12-x^2
(12)+xy=18
xy=6

Attempt 2:
xy=6
x=y/6
y^2/36+(y/6)y+y^2=18
43/36y^2=18
y ≠ root(6) <- should be the answer

Edit:
Just realized you can't plug the modified equation back into its original self

I plugged y=6/x into the circle instead and got:
x^2+36/x^2=12
now I have x^4+36=12x^2
x^4-12x^2+36=0

 

[Mark44]

Homework Statement:: Find the intersection/system of equation between a parabola and circle
Relevant Equations:: x^2+xy+y^2=18; x^2+y^2=12

Given:
x^2+xy+y^2=18
x^2+y^2=12

Attempt:
(x^2+y^2)+xy=18
12+xy=18
xy=6

y^2=12-x^2
(12)+xy=18
xy=6

Attempt 2:
xy=6
x=y/6
y^2/36+(y/6)y+y^2=18
43/36y^2=18
y ≠ root(6) <- should be the answer

Edit:
Just realized you can't plug the modified equation back into its original self

I plugged y=6/x into the circle instead and got:
x^2+36/x^2=12
now I have x^4+36=12x^2
x^4-12x^2+36=0

Your last equation is quadratic in form. Let $u = x^2$ to get an actual quadratic, and solve for u, then substititute back to get values for x.

 

[SammyS]

Attempt 2:
xy=6
x=y/6
...

Edit:
Just realized you can't plug the modified equation back into its original self
...

No. That wasn't what caused your difficulty.

If you solve
$xy=6$ for $x$, you get $x=\dfrac {6}{y}$ rather than $x=\dfrac {y}{6}$. Plugging that into either equation gives the correct solution.