Question

what are the co - vertices of the ellipse $\frac{x^{2}}{36}+\frac{(y - 6)^{2}}{4}=1$?
write your answer in simplified, rationalized form.
(, ) and (, )

Explanation:

Step1: Identify the form of the ellipse equation

The standard form of an ellipse is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\). For the given equation \(\frac{x^{2}}{36}+\frac{(y - 6)^{2}}{4}=1\), we have \(h = 0\), \(k=6\), \(a^{2}=36\) (so \(a = 6\)), \(b^{2}=4\) (so \(b = 2\)). Since \(a>b\), this is a horizontal - major - axis ellipse.

Step2: Recall the formula for co - vertices

For a horizontal - major - axis ellipse \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\), the co - vertices are \((h,k + b)\) and \((h,k - b)\)

Step3: Calculate the co - vertices

Substitute \(h = 0\), \(k = 6\), \(b=2\) into the formula.
For the first co - vertex: \(x=h = 0\), \(y=k + b=6 + 2=8\)
For the second co - vertex: \(x=h = 0\), \(y=k - b=6-2 = 4\)

Answer:

\((0,8)\) and \((0,4)\)