Question

find the center of this ellipse.
$x^2 + 4y^2 - 10x - 40y + 121 = 0$
center : (? , )

Explanation:

Step1: Group x and y terms

Group the \(x\)-terms and \(y\)-terms together: \((x^{2}-10x)+(4y^{2}-40y)= - 121\)

Step2: Factor out coefficients of squared terms

Factor out the coefficient of \(y^{2}\) from the \(y\)-terms: \((x^{2}-10x)+4(y^{2}-10y)= - 121\)

Step3: Complete the square for x

For the \(x\)-terms: \(x^{2}-10x=(x - 5)^{2}-25\) (since \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a=x\), \(2ab = 10x\) so \(b = 5\))

Step4: Complete the square for y

For the \(y\)-terms: \(y^{2}-10y=(y - 5)^{2}-25\), so \(4(y^{2}-10y)=4((y - 5)^{2}-25)=4(y - 5)^{2}-100\)

Step5: Substitute back and simplify

Substitute the completed square forms back into the equation:
\((x - 5)^{2}-25+4(y - 5)^{2}-100=-121\)
Simplify the left - hand side: \((x - 5)^{2}+4(y - 5)^{2}-125=-121\)
Then, \((x - 5)^{2}+4(y - 5)^{2}=4\)
Divide both sides by 4: \(\frac{(x - 5)^{2}}{4}+\frac{(y - 5)^{2}}{1}=1\)

The standard form of an ellipse is \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1\), where \((h,k)\) is the center. Here \(h = 5\) and \(k = 5\)

Answer:

\((5,5)\)