Question

question 14 of 20
which of the following expresses the coordinates of the foci of the conic section shown below?
\\(\frac{(x + 2)^2}{49}+\frac{(y - 1)^2}{25}=1\\)
a. \\((-2\pm2\sqrt{6},1)\\)
b. \\((-2,1\pm\sqrt{74})\\)
c. \\((-2,1\pm2\sqrt{6})\\)
d. \\((-2\pm\sqrt{74},1)\\)

Explanation:

Step1: Identify the form of the ellipse

The equation $\frac{(x + 2)^2}{49}+\frac{(y - 1)^2}{25}=1$ is of the form $\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1$ where $h=-2,k = 1,a^2 = 49$ (so $a = 7$), $b^2=25$ (so $b = 5$).

Step2: Calculate $c$

For an ellipse, $c^2=a^2 - b^2$. Substitute $a^2 = 49$ and $b^2 = 25$ into the formula: $c^2=49 - 25=24$, so $c = 2\sqrt{6}$.

Step3: Determine the foci

Since the major - axis is parallel to the $x$ - axis (because $a^2>b^2$), the foci of the ellipse $\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1$ are given by the coordinates $(h\pm c,k)$. Substitute $h=-2,k = 1,c = 2\sqrt{6}$ into the formula, we get the foci as $(-2\pm2\sqrt{6},1)$.

Answer:

A. $(-2\pm2\sqrt{6},1)$