Question

what are the vertices of the ellipse (5x^{2}+3y^{2}-10x - 12y-58 = 0)?
write your answer in simplified, rationalized form.
((quad,quad)) and ((quad,quad))

Explanation:

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

Group \(x\) - terms and \(y\) - terms:

$$
LATEXBLOCK0
$$

For the \(x\) - part: \(x^{2}-2x=(x - 1)^{2}-1\). For the \(y\) - part: \(y^{2}-4y=(y - 2)^{2}-4\)

$$
LATEXBLOCK1
$$

Divide by \(75\) to get the standard form \(\frac{(x - 1)^{2}}{15}+\frac{(y - 2)^{2}}{25}=1\)

Step2: Identify the center, \(a\) and \(b\)

The standard form of an ellipse is \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) (\(a>b>0\)), where \((h,k)\) is the center. Here \(h = 1,k = 2,a = 5,b=\sqrt{15}\)
The vertices of the ellipse \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) are \((h,k\pm a)\)

Step3: Calculate the vertices

Substitute \(h = 1,k = 2,a = 5\)
The vertices are \((1,2 + 5)\) and \((1,2-5)\)

Answer:

\((1,7)\) and \((1,-3)\)