Question

find the center and foci of the ellipse.

25x² + 4y² - 150x - 64y + 381 = 0

center = (?, )

foci = (, ± √)

Explanation:

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

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

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For the \(x\) - part: \(x^{2}-6x=(x - 3)^{2}-9\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a = x\), \(b = 3\))
For the \(y\) - part: \(y^{2}-16y=(y - 8)^{2}-64\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a = y\), \(b = 8\))

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Step2: Identify the center

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.
Comparing \(\frac{(x - 3)^{2}}{4}+\frac{(y - 8)^{2}}{25}=1\) with \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\), we get \(h = 3\), \(k = 8\)

Step3: Calculate \(c\) (for foci)

We know the relationship \(c^{2}=a^{2}-b^{2}\). Here \(a^{2}=25\), \(b^{2}=4\), so \(c^{2}=25 - 4=21\), \(c=\sqrt{21}\)

The foci of the ellipse \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) (\(a>b>0\)) are \((h,k\pm c)\)

Answer:

Center \(=(3,8)\)
Foci \(=(3,8\pm\sqrt{21})\)