Question

what are the co - vertices of the ellipse \\( \frac{(x - 3)^{2}}{36}+\frac{(y - 1)^{2}}{7}=1 \\)?
write your answer in simplified, rationalized form.
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Explanation:

Step1: Identify the standard form of the ellipse

The standard form of an ellipse is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (for \(a>b\)), where \((h,k)\) is the center. For the given ellipse \(\frac{(x - 3)^2}{36}+\frac{(y - 1)^2}{7}=1\), we have \(h = 3\), \(k = 1\), \(a^2=36\) (so \(a = 6\)), \(b^2 = 7\) (so \(b=\sqrt{7}\)).

Step2: Recall the formula for co - vertices

For an ellipse of the form \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (\(a>b\)), the co - vertices are \((h,k + b)\) and \((h,k - b)\)

Substitute \(h = 3\), \(k = 1\), \(b=\sqrt{7}\) into the formula.

For \((h,k + b)\): \(x=3\), \(y=1+\sqrt{7}\)

For \((h,k - b)\): \(x = 3\), \(y=1-\sqrt{7}\)

Answer:

\((3,1+\sqrt{7})\) and \((3,1 - \sqrt{7})\)