Question

write the equation of the ellipse $9x^{2}+36y^{2}-90x - 288y + 477 = 0$ in standard form

$\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1$,

where:
$h=$
$k=$
$a=$
$b=$
question help: video 1 video 2 post to forum

Explanation:

Step1: Group \(x\) and \(y\) terms

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Step2: Complete the square for \(x\) and \(y\)

For \(x\): \((x - 5)^{2}=x^{2}-10x + 25\), so \(x^{2}-10x=(x - 5)^{2}-25\)
For \(y\): \((y - 4)^{2}=y^{2}-8y + 16\), so \(y^{2}-8y=(y - 4)^{2}-16\)

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Step3: Divide by 324 to get the standard form

$$ \frac{(x - 5)^{2}}{36}+\frac{(y - 4)^{2}}{9}=1 $$

Answer:

\(h = 5\), \(k = 4\), \(a = 6\), \(b = 3\)