Conics Equation and Circle Problem: Solving for Unknown Variables

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Hysteria X
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Homework Statement



The equation ##x^2y^2-2xy^2-3y^2-4x^2y+8xy+12y=0## represents??

Homework Equations



circle: ##x^2 +y^2 = a^2##

The Attempt at a Solution


i know this has something to do with seperating out the variables but i don't seem to get the req equation
 
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Hysteria X said:

Homework Statement



The equation ##x^2y^@-2xy-3y^2-4x^2y+8xy+12y=0## represents??

Homework Equations



circle: ##x^2 +y^2 = a^2##

The Attempt at a Solution


i know this has something to do with seperating out the variables but i don't seem to get the req equation

Show us what you've tried.
 
Mark44 said:
Show us what you've tried.
i divided the whole term by y^2 and i separated the y and x terms on both sides of the equation then i think the next step would be to convert into factors but how am i supposed to do that why y would be in the denominator in rhs??
 
Dick said:
You want to try to factor it somehow. Why are there two xy terms in your expression? Check for typos.

sorry its ##xy^2##
 
Dick said:
Ok, then start trying to factor it. You can pull a y out right away.

##x^2y^2−2xy^2−3y^2−4x^2y+8xy+12y=0##
##y^2(x^2-2x-3)-4y(x^2-2x-3)=0##
##y-4=0##
##y=4##? what conic is that? is it a straight line :confused:
 
Hysteria X said:
##x^2y^2−2xy^2−3y^2−4x^2y+8xy+12y=0##
##y^2(x^2-2x-3)-4y(x^2-2x-3)=0##
##y-4=0##
##y=4##? what conic is that? is it a straight line :confused:

Yes, it's a line. It can happen. xy=1 is a hyperbola. xy=0 is two lines. That's a 'degenerate conic'. But actually since your equation is 4th degree, there's not necessarily any reason to expect it to be a conic. But y=4 isn't the whole story.
 
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Hysteria X said:
##x^2y^2−2xy^2−3y^2−4x^2y+8xy+12y=0##
##y^2(x^2-2x-3)-4y(x^2-2x-3)=0##
You skipped some steps here. Write the equation above as a product instead of a difference. In the two terms above there is a common factor: x2 - 2x - 3.
Hysteria X said:
##y-4=0##
##y=4##? what conic is that? is it a straight line :confused: