Converting standard form to standard form of an ellipse

AI Thread Summary
To convert the equation x^2 + 4y^2 - 8y - 6x + 9 into the standard form of an ellipse, it is essential to rearrange it to include an equals sign, resulting in x^2 + 4y^2 - 8y - 6x + 9 = 0. The correct standard form should be expressed as (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h, k) is the center of the ellipse. The discussion highlights confusion regarding the presence of a constant term, specifically -4, which needs clarification in the context of the standard form. Ultimately, the user successfully resolves their misunderstanding and appreciates the guidance provided. Understanding the proper format is crucial for identifying the foci and center of the ellipse.
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Homework Statement


x^2+4y^2-8y-6x+9 convert this equation to the standard form of an ellipse and identify the foci, the middle point,

Homework Equations


x^2/b^2+y^2/b^2


The Attempt at a Solution



Ellipse-----> My standard form of this ellipse ((x-3)^2/(2)^2)+((y-2)^2/(1^2))-4
I worked at this as the form of the ellipse but i just do not understand what i am doing wrong because the -4 is there.

 
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x^2+4y^2-8y-6x+9 isn't an 'equation'. There's no '=' sign. Neither is x^2/b^2+y^2/b^2. If you write the standard form as x^2/b^2+y^2/b^2=1 and the form given as x^2+4y^2-8y-6x+9=0, it might be clearer what you should do with the -4.
 
Thanks so much! Yeah i left out some pretty important stuff.. I got the answer now. Appreciate it!
 
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