Question

question 3 of 10
which of the following correctly represents the coordinates of the foci of the ellipse shown below?
$\frac{(x - 7)^2}{4}+\frac{(y + 3)^2}{16}=1$
a. $(7,-3pm2sqrt{5})$
b. $(7pm2sqrt{5},-3)$
c. $(7,-3pm2sqrt{3})$
d. $(7pm2sqrt{3},-3)$

Explanation:

Step1: Identify the form of the ellipse equation

The standard - form of an ellipse centered at \((h,k)\) is \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (\(a>b>0\)). Given \(\frac{(x - 7)^2}{4}+\frac{(y + 3)^2}{16}=1\), we have \(h = 7\), \(k=-3\), \(a^2 = 16\) (so \(a = 4\)), and \(b^2 = 4\) (so \(b = 2\)).

Step2: Calculate the value of \(c\)

The relationship for an ellipse is \(c^2=a^2 - b^2\). Substitute \(a = 4\) and \(b = 2\) into the formula: \(c^2=16 - 4=12\), so \(c = 2\sqrt{3}\).

Step3: Determine the foci coordinates

For an ellipse of the form \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\), the foci are located at \((h,k\pm c)\). Substitute \(h = 7\), \(k=-3\), and \(c = 2\sqrt{3}\) into the formula, we get the foci coordinates \((7,-3\pm2\sqrt{3})\).

Answer:

C. \((7,-3\pm2\sqrt{3})\)