Unable to find the intersection between a circle and ellipse

  • Thread starter lotur512
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  • #1
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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 cant 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
 
Last edited:

Answers and Replies

  • #2
35,287
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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 cant 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.
 
  • #3
SammyS
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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.
 

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