Conics Equation and Circle Problem: Solving for Unknown Variables

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Homework Help Overview

The problem involves analyzing the equation ##x^2y^2-2xy^2-3y^2-4x^2y+8xy+12y=0## to determine the type of conic section it represents. Participants are exploring the relationship between the given equation and known forms of conic sections, particularly circles.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss factoring the equation and separating variables. There are questions about the presence of multiple xy terms and potential typos in the equation. Some participants express uncertainty about the implications of their manipulations and the resulting forms.

Discussion Status

There is ongoing exploration of the equation's structure, with some participants attempting to factor it and others questioning the nature of the resulting expressions. Guidance has been offered regarding factoring and checking for errors, but no consensus has been reached on the overall interpretation of the equation.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the depth of their explorations. There is also a focus on understanding the implications of the equation's degree and its relation to conic sections.

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
 
Last edited:
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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.
 
You want to try to factor it somehow. Why are there two xy terms in your expression? Check for typos.
 
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##
 
Hysteria X said:
sorry its ##xy^2##

Ok, then start trying to factor it. You can pull a y out right away.
 
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:
 

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