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The equation of an ellipse is given below. ((x-1)^(2))/(45)+(y^(2))/(54)=1 What are the foci of this ellipse? Choose 1 answer: (A) (1,3) and (1,-3) (B) (-1,3) and (-1,-3) (C) (4,0) and (-2,0) (D) (3,-1) and (-3,-1)

The equation of an ellipse is given below.\newline(x−1)245+y254=1 \frac{(x-1)^{2}}{45}+\frac{y^{2}}{54}=1 \newlineWhat are the foci of this ellipse?\newlineChoose 11 answer:\newline(A) (1,3) (1,3) and (1,−3) (1,-3) \newline(B) (−1,3) (-1,3) and (−1,−3) (-1,-3) \newline(C) (4,0) (4,0) and (−2,0) (-2,0) \newline(D) (3,−1) (3,-1) and (−3,−1) (-3,-1)

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Q. The equation of an ellipse is given below.\newline(x−1)245+y254=1 \frac{(x-1)^{2}}{45}+\frac{y^{2}}{54}=1 \newlineWhat are the foci of this ellipse?\newlineChoose 11 answer:\newline(A) (1,3) (1,3) and (1,−3) (1,-3) \newline(B) (−1,3) (-1,3) and (−1,−3) (-1,-3) \newline(C) (4,0) (4,0) and (−2,0) (-2,0) \newline(D) (3,−1) (3,-1) and (−3,−1) (-3,-1)
  1. Identify standard form: Identify the standard form of the ellipse equation.\newlineThe given equation is already in the standard form of an ellipse, which 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 semi-major axis, and bb is the semi-minor axis.
  2. Determine values of a2a^2, b2b^2, and center: Determine the values of a2a^2, b2b^2, and the center (h,k)(h, k). From the given equation (x−1)245+y254=1\frac{(x-1)^2}{45} + \frac{y^2}{54} = 1, we can see that a2=45a^2 = 45, b2=54b^2 = 54, and the center (h,k)=(1,0)(h, k) = (1, 0).
  3. Identify major axis: Identify which axis is the major axis. Since b2>a2b^2 > a^2, the major axis is along the y-axis.
  4. Calculate distance cc: Calculate the distance cc from the center to the foci.\newlineThe distance cc is found using the equation c2=b2−a2c^2 = b^2 - a^2. Let's calculate it.\newlinec2=54−45c^2 = 54 - 45\newlinec2=9c^2 = 9\newlinec=9c = \sqrt{9}\newlinec=3c = 3
  5. Determine coordinates of foci: Determine the coordinates of the foci.\newlineSince the major axis is along the y-axis, the foci will be at (h,k±c)(h, k \pm c). Substituting the values we have:\newlineFoci = (1,0±3)(1, 0 \pm 3)\newlineThis gives us the two foci at (1,3)(1, 3) and (1,−3)(1, -3).

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