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: \(x^{2}-6x=(x - 3)^{2}-9\). For the \(y\) - part: \(y^{2}-16y=(y - 8)^{2}-64\)

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Step2: Identify the center

The standard form of an ellipse is \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) (\(a>b>0\)). Here \(h = 3,k = 8\), so the center \((h,k)=(3,8)\)

Step3: Calculate \(c\) (distance from center to foci)

We know that for an ellipse \(c^{2}=a^{2}-b^{2}\). Given \(a^{2}=25\), \(b^{2}=4\), then \(c^{2}=25 - 4=21\), \(c=\sqrt{21}\)

Step4: Find the foci

Since the major axis is along the \(y\) - axis (because \(a^{2}=25\) is under the \(y\) - term), the foci are \((h,k\pm c)\)

Answer:

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