Question

what are the co - vertices of the ellipse ( x^{2}+4y^{2}-32y + 48 = 0 )?
write your answer in simplified, rationalized form.
( (quad,quad) ) and ( (quad,quad) )

Explanation:

Step1: Complete the square for the \(y\) terms

Given the equation \(x^{2}+4y^{2}-32y + 48=0\).
First, factor out the coefficient of \(y^{2}\) from the \(y\) - terms:
\(x^{2}+4(y^{2}-8y)+48 = 0\).
Complete the square for \(y^{2}-8y\): \((y - 4)^{2}=y^{2}-8y + 16\).
So, \(x^{2}+4((y - 4)^{2}-16)+48 = 0\).
Expand: \(x^{2}+4(y - 4)^{2}-64 + 48=0\).
Simplify to get the standard form of the ellipse equation \(\frac{x^{2}}{16}+\frac{(y - 4)^{2}}{4}=1\).
The standard form of an ellipse is \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1\) (\(a>b\)), where \((h,k)\) is the center, \(a\) is the semi - major axis, and \(b\) is the semi - minor axis. Here, \(h = 0,k = 4,a = 4,b = 2\).

Step2: Find the co - vertices

For the ellipse \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1\) (\(a>b\)), the co - vertices are \((h\pm b,k)\).
Substituting \(h = 0,k = 4,b = 2\), we get the co - vertices \((0 + 2,4)\) and \((0-2,4)\).

Answer:

\((2,4)\) and \((-2,4)\)