Question

what are the coordinates of the center of the ellipse shown below?\\(\frac{(x - 7)^2}{4} + \frac{(y + 3)^2}{16} = 1\\)\
a, \\((-7, 3)\\)\
b. \\((2, 4)\\)\
c. \\((7, -3)\\)\
d. \\((4, 16)\\)

Explanation:

Step1: Recall the standard form of an ellipse equation

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

Step2: Rewrite the given equation in the standard - form

The given equation is \(\frac{(x - 7)^{2}}{4}+\frac{(y+3)^{2}}{16}=1\). We can rewrite \(y + 3\) as \(y-(-3)\).
Comparing \(\frac{(x - 7)^{2}}{4}+\frac{(y-(-3))^{2}}{16}=1\) with \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}} = 1\), we get \(h = 7\) and \(k=-3\).

Answer:

C. \((7,-3)\)