Question

find the center and foci of the ellipse: 16x² + 25y² - 64x - 50y - 311 = 0 center = (?, ) foci = (, ) and (, )

Explanation:

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

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

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

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Divide by \(400\) to get the standard form \(\frac{(x - 2)^{2}}{25}+\frac{(y - 1)^{2}}{16}=1\)

Step2: Identify the center \((h,k)\)

For the ellipse \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}} = 1\) (\(a>b>0\)), the center is \((h,k)\). Here \(h = 2\), \(k = 1\)

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

We know that \(c^{2}=a^{2}-b^{2}\). Since \(a^{2}=25\), \(b^{2}=16\), then \(c^{2}=25 - 16=9\), so \(c = 3\)

Step4: Find the foci

The foci of the ellipse \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1\) (\(a>b>0\)) are \((h\pm c,k)\)
Substitute \(h = 2\), \(k = 1\), \(c = 3\)
The foci are \((2 + 3,1)=(5,1)\) and \((2-3,1)=(-1,1)\)

Answer:

Center \(=(2,1)\)
Foci \(=(5,1)\) and \((-1,1)\)