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The equation of an ellipse is given below. ((x-3)^(2))/(25)+((y+1)^(2))/(169)=1 What are the foci of this ellipse? Choose 1 answer: (A) (3,-13) and (3,11) (B) (9,1) and (-15,1) (c) (-12,-1) and (15,-1) (D) (-3,13) and (-3,-11)

The equation of an ellipse is given below.\newline(x−3)225+(y+1)2169=1 \frac{(x-3)^{2}}{25}+\frac{(y+1)^{2}}{169}=1 \newlineWhat are the foci of this ellipse?\newlineChoose 11 answer:\newline(A) (3,−13) (3,-13) and (3,11) (3,11) \newline(B) (9,1) (9,1) and (−15,1) (-15,1) \newline(C) (−12,−1) (-12,-1) and (15,−1) (15,-1) \newline(D) (−3,13) (-3,13) and (−3,−11) (-3,-11)

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Q. The equation of an ellipse is given below.\newline(x−3)225+(y+1)2169=1 \frac{(x-3)^{2}}{25}+\frac{(y+1)^{2}}{169}=1 \newlineWhat are the foci of this ellipse?\newlineChoose 11 answer:\newline(A) (3,−13) (3,-13) and (3,11) (3,11) \newline(B) (9,1) (9,1) and (−15,1) (-15,1) \newline(C) (−12,−1) (-12,-1) and (15,−1) (15,-1) \newline(D) (−3,13) (-3,13) and (−3,−11) (-3,-11)
  1. Identify center and axes lengths: Identify the center and lengths of the semi-major and semi-minor axes.\newlineThe standard form of an ellipse is (x−h)2/a2+(y−k)2/b2=1(x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k)(h, k) is the center of the ellipse, aa is the length of the semi-major axis, and bb is the length of the semi-minor axis. For the given ellipse, h=3h = 3, k=−1k = -1, a2=25a^2 = 25, and b2=169b^2 = 169. Therefore, a=5a = 5 and b=13b = 13.
  2. Determine major axis: Determine which axis is the major axis.\newlineSince b>ab > a, the major axis is along the y-axis. This means that the foci will be vertically aligned with the center of the ellipse.
  3. Calculate distance to foci: Calculate the distance cc from the center to each focus. The distance cc is found using the equation c2=b2−a2c^2 = b^2 - a^2. Plugging in the values, we get c2=169−25=144c^2 = 169 - 25 = 144. Taking the square root gives us c=12c = 12.
  4. Find foci coordinates: Find the coordinates of the foci.\newlineThe foci are located at (h,k±c)(h, k \pm c) since the major axis is vertical. Substituting the values, we get the foci at (3,−1±12)(3, -1 \pm 12). This gives us the two points (3,−1+12)(3, -1 + 12) and (3,−1−12)(3, -1 - 12), which simplifies to (3,11)(3, 11) and (3,−13)(3, -13).

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