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The equation of an ellipse is given below. ((x-4)^(2))/(9)+((y+6)^(2))/(4)=1 What are the foci of this ellipse? Choose 1 answer: (A) (4+sqrt5,-6) and (4-sqrt5,-6) (B) (13,-6) and (-5,-6) (c) (4,3) and (4,-15) (D) (4,-6+sqrt5) and (4,-6-sqrt5)

The equation of an ellipse is given below.\newline(x−4)29+(y+6)24=1 \frac{(x-4)^{2}}{9}+\frac{(y+6)^{2}}{4}=1 \newlineWhat are the foci of this ellipse?\newlineChoose 11 answer:\newline(A) (4+5,−6) (4+\sqrt{5},-6) and (4−5,−6) (4-\sqrt{5},-6) \newline(B) (13,−6) (13,-6) and (−5,−6) (-5,-6) \newline(C) (4,3) (4,3) and (4,−15) (4,-15) \newline(D) (4,−6+5) (4,-6+\sqrt{5}) and (4,−6−5) (4,-6-\sqrt{5})

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Q. The equation of an ellipse is given below.\newline(x−4)29+(y+6)24=1 \frac{(x-4)^{2}}{9}+\frac{(y+6)^{2}}{4}=1 \newlineWhat are the foci of this ellipse?\newlineChoose 11 answer:\newline(A) (4+5,−6) (4+\sqrt{5},-6) and (4−5,−6) (4-\sqrt{5},-6) \newline(B) (13,−6) (13,-6) and (−5,−6) (-5,-6) \newline(C) (4,3) (4,3) and (4,−15) (4,-15) \newline(D) (4,−6+5) (4,-6+\sqrt{5}) and (4,−6−5) (4,-6-\sqrt{5})
  1. Given equation of the ellipse: Given equation of the ellipse: (x−4)29+(y+6)24=1\frac{(x-4)^{2}}{9}+\frac{(y+6)^{2}}{4}=1. Identify the center (h,k)(h, k), the lengths of the semi-major axis (a)(a), and the semi-minor axis (b)(b). The center is at (h,k)=(4,−6)(h, k) = (4, -6). The length of the semi-major axis is the square root of the larger denominator, so a=9=3a = \sqrt{9} = 3. The length of the semi-minor axis is the square root of the smaller denominator, so b=4=2b = \sqrt{4} = 2.
  2. Identify the center, lengths of the semi-major axis, and semi-minor axis: Calculate the distance cc from the center to the foci using the formula c=a2−b2c = \sqrt{a^2 - b^2}. Here, a=3a = 3 and b=2b = 2. c=32−22=9−4=5c = \sqrt{3^2 - 2^2} = \sqrt{9 - 4} = \sqrt{5}.
  3. Calculate the distance cc from the center to the foci: Determine the coordinates of the foci.\newlineSince the larger denominator is under the (x−4)2(x-4)^2 term, the foci are horizontal from the center.\newlineThe foci are at (h±c,k)=(4±5,−6)(h \pm c, k) = (4 \pm \sqrt{5}, -6).
  4. Determine the coordinates of the foci: Write the final coordinates of the foci.\newlineThe foci are at (4+5,−6)(4 + \sqrt{5}, -6) and (4−5,−6)(4 - \sqrt{5}, -6).

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