Also help with conicsmainly ellipses?

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The discussion focuses on converting the equation x^2 + 4y^2 - 2x - 32y = 0 into standard form for an ellipse. The solution involves completing the square for both x and y terms, resulting in the equation (x-1)^2 + 4(y-4)^2 = 65. To express it in standard form, the equation is then divided by 65, yielding (x-1)^2/65 + (y-4)^2/(65/4) = 1. Participants confirm that the steps taken are correct and the final form is appropriate for an ellipse. Overall, the conversion process and the resulting standard form are validated by peers in the discussion.
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Homework Statement



x^2 + 4y^2 - 2x - 32y = 0
Get into standard form

Homework Equations





The Attempt at a Solution


x^2 + 4y^2 - 2x - 32y = 0
Complete the square two times.
(x^2 - 2x + 1) +4(y^2 - 8y +16 )=0 + 4(16) + 1
Simplfy..
(x-1)^2 + 4(y-4)^2 =65
Try to get the equation to be equal to one.. so divide by 65.
(x-1)^2/(65) + (y-4)^/(65/4) = 1
Did I do this right? This just looks really strange...
 
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It doesn't look strange to me. It looks correct.
 
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