Question

find the center and foci of the ellipse.
64x² + 100y² + 768x - 600y - 3196 = 0
center = (?, )
foci = (, ) and (, )

Explanation:

Step1: Group and factor terms

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

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Step2: Complete the square

For the \(x\) - part: \((x + 6)^{2}=x^{2}+12x + 36\), so \(64(x^{2}+12x)=64((x + 6)^{2}-36)\)
For the \(y\) - part: \((y - 3)^{2}=y^{2}-6y+9\), so \(100(y^{2}-6y)=100((y - 3)^{2}-9)\)

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Step3: Write in standard form

Divide by \(6400\): \(\frac{(x + 6)^{2}}{100}+\frac{(y - 3)^{2}}{64}=1\)

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

Here \(h=-6,k = 3,a = 10,b = 8\), \(c=\sqrt{100 - 64}=\sqrt{36}=6\)

Answer:

Center \(=(-6,3)\)
Foci \(=(-6 + 6,3)=(0,3)\) and \((-6-6,3)=(-12,3)\)