Question

find the center of this ellipse.
x² + 4y² - 10x - 40y + 121 = 0
center : ( ?, )

Explanation:

Step1: Rearrange and complete the square for x - terms

Group the x - terms and y - terms: $(x^{2}-10x)+4(y^{2}-10y)+121 = 0$.
For the x - terms: $x^{2}-10x=(x - 5)^{2}-25$.

Step2: Complete the square for y - terms

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

Step3: Rewrite the equation of the ellipse

Substitute the completed - square expressions back into the original equation:
$(x - 5)^{2}-25+4(y - 5)^{2}-100 + 121=0$.
Simplify the equation: $(x - 5)^{2}+4(y - 5)^{2}-4 = 0$, then $(x - 5)^{2}+4(y - 5)^{2}=4$, or $\frac{(x - 5)^{2}}{4}+\frac{(y - 5)^{2}}{1}=1$.

Answer:

$(5,5)$