Find Equation for Hyperbola or Ellipse

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SUMMARY

The discussion focuses on finding the equation of a hyperbola or ellipse from the given quadratic equation: 6x² + 8y² + 32y - 16 = 0. The correct approach involves rearranging the equation to isolate the terms and completing the square for the y terms. The final form should resemble \(\frac{x^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), indicating the standard form of a conic section. The user initially made an error in their calculations, leading to an incorrect equation.

PREREQUISITES
  • Understanding of quadratic equations and their standard forms
  • Knowledge of completing the square technique
  • Familiarity with conic sections, specifically hyperbolas and ellipses
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the method of completing the square in quadratic equations
  • Learn about the standard forms of conic sections, including hyperbolas and ellipses
  • Practice converting quadratic equations into conic section forms
  • Explore examples of identifying conic sections from their equations
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Students studying algebra, particularly those focusing on conic sections, as well as educators looking for examples of solving quadratic equations related to hyperbolas and ellipses.

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Homework Statement




6x2 + 8y2 + 32y - 16 = 0

Homework Equations





The Attempt at a Solution



I think I made a mistake.

This is how far I got

4(x-4)^+3(y+9)^=120


I made a mistake. Can someone delete this thread?

What did I do wrong?
 
Last edited:
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realism877 said:

Homework Statement




6x2 + 8y2 + 32y - 16 = 0

Homework Equations





The Attempt at a Solution



I think I made a mistake.

This is how far I got

4(x-4)^+3(y+9)^=120


I made a mistake. Can someone delete this thread?

What did I do wrong?

I can't tell what you did wrong since you didn't show your work, but the equation you ended with isn't right.

6x2 + 8y2 + 32y - 16 = 0
<==> 6x2 + 8y2 + 32y = 16
Now complete the square in the y terms so that the equation becomes
\frac{x^2}{a^2} + \frac{(y - k)^2}{b^2} = 1

Since there is no first-degree x term in the original equation, don't try to complete the square in the x terms.
 

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