Question

what is the center of the ellipse $13x^{2}+4y^{2}-52 = 0$? write your answer in simplified, rationalized form.

Explanation:

Step1: Rewrite the equation in standard form

The general equation of an ellipse is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\), where \((h,k)\) is the center.
Given \(13x^{2}+4y^{2}-52 = 0\), we can rewrite it as \(\frac{x^{2}}{4}+\frac{y^{2}}{13}=1\) (by dividing each term by \(52\): \(\frac{13x^{2}}{52}+\frac{4y^{2}}{52}-\frac{52}{52}=0\) simplifies to \(\frac{x^{2}}{4}+\frac{y^{2}}{13}=1\)).
In the form \(\frac{(x - 0)^2}{4}+\frac{(y - 0)^2}{13}=1\), here \(h = 0\) and \(k = 0\).

Answer:

\((0,0)\)