MHB Graphing Ellipses: How to Change Formats

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To graph the ellipse given by the equation (x+2)²/5 + 2(y-1)² = 1, it needs to be rearranged into the standard form for ellipses. The correct form is (x-x₀)²/a² + (y-y₀)²/b² = 1, where the center is at (-2, 1). The coefficient of 2 in front of (y-1)² can be adjusted by rewriting it as (y-1)²/(1/2) to fit the standard format. This allows for proper graphing of the ellipse. Understanding these transformations is essential for accurately representing the ellipse on a graph.
veronica1999
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How do I graph this ellipse?

It doesn't seem to be in the right form.

(x+2)^2 /5 + 2 (y-1)^2 = 1

I don't know what to do with the 2 in front of the (y-1)^2

Doesn't an ellipse have to be x^2/a^2 + y^2/b^2 = 1
 
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veronica1999 said:
How do I graph this ellipse?

It doesn't seem to be in the right form.

(x+2)^2 /5 + 2 (y-1)^2 = 1

I don't know what to do with the 2 in front of the (y-1)^2

Doesn't an ellipse have to be x^2/a^2 + y^2/b^2 = 1

Hi veronica1999, :)

Yes, an ellipse has its equation as, \(\displaystyle\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) if its major and minor axes coincides with the \(x\) and \(y\) axes of the Cartesian coordinate system. In this case the center point of the ellipse is at the origin. However a ellipse with its center point at, \((x_{0},y_{0})\) has an equation of the form,

\[\frac{(x-x_{0})^2}{a^2} + \frac{(y-y_{0})^2}{b^2} = 1\]

In your case the ellipse is centered at, \((-2,1)\). Now you should be able to draw your ellipse. :)

Kind Regards,
Sudharaka.
 
Thanks.
But I am still not sure what to do with the 2 in front of the (y-1)^2.
Could it have been a mistype meaning (y-1)^2/2 instead of 2(y-1)^2?
 
veronica1999 said:
Thanks.
But I am still not sure what to do with the 2 in front of the (y-1)^2.
Could it have been a mistype meaning (y-1)^2/2 instead of 2(y-1)^2?

Of course not. You can rearrange the equation by taking that \(2\) to the denominator like this,

\[\frac{(x+2)^2}{5} + \frac{(y-1)^2}{\frac{1}{2}} = 1\]

Is it clear to you now? :)
 
Yes!;)

Thank you!
 
veronica1999 said:
Yes!;)

Thank you!

You are welcome. I am glad to be of any help. :)
 
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