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What are the foci of the hyperbola represented by the equation (x^(2))/(16)-(y^(2))/(25)=1? Choose 1 answer: (A) (3,0) and (-3,0) (B) (sqrt41,0) and (-sqrt41,0) (c) (0,3) and (0,-3) (D) (0,sqrt41) and (0,-sqrt41)

What are the foci of the hyperbola represented by the equation x216−y225=1? \frac{x^{2}}{16}-\frac{y^{2}}{25}=1 ? \newlineChoose 11 answer:\newline(A) (3,0) (3,0) and (−3,0) (-3,0) \newline(B) (41,0) (\sqrt{41}, 0) and (−41,0) (-\sqrt{41}, 0) \newline(C) (0,3) (0,3) and (0,−3) (0,-3) \newline(D) (0,41) (0, \sqrt{41}) and (0,−41) (0,-\sqrt{41})

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Q. What are the foci of the hyperbola represented by the equation x216−y225=1? \frac{x^{2}}{16}-\frac{y^{2}}{25}=1 ? \newlineChoose 11 answer:\newline(A) (3,0) (3,0) and (−3,0) (-3,0) \newline(B) (41,0) (\sqrt{41}, 0) and (−41,0) (-\sqrt{41}, 0) \newline(C) (0,3) (0,3) and (0,−3) (0,-3) \newline(D) (0,41) (0, \sqrt{41}) and (0,−41) (0,-\sqrt{41})
  1. Equation of the hyperbola: The given equation is of a hyperbola in the standard form (x2a2)−(y2b2)=1(\frac{x^2}{a^2}) - (\frac{y^2}{b^2}) = 1, where a2a^2 is under the x2x^2 term and b2b^2 is under the y2y^2 term. For a hyperbola centered at the origin with a horizontal transverse axis, the foci are located at (±c,0)(\pm c, 0), where cc is found using the equation c2=a2+b2c^2 = a^2 + b^2.
  2. Identifying a2a^2 and b2b^2: Identify the values of a2a^2 and b2b^2 from the given equation. Here, a2=16a^2 = 16 and b2=25b^2 = 25.
  3. Calculating c2c^2: Calculate the value of c2c^2 using the equation c2=a2+b2c^2 = a^2 + b^2. Substituting the identified values gives us c2=16+25c^2 = 16 + 25.
  4. Finding the value of c: Perform the addition to find c2c^2. So, c2=41c^2 = 41.
  5. Locating the foci: Find the value of cc by taking the square root of c2c^2. Therefore, c=41c = \sqrt{41}.
  6. Locating the foci: Find the value of cc by taking the square root of c2c^2. Therefore, c=41c = \sqrt{41}.The foci of the hyperbola are at (±c,0)(\pm c, 0). Substituting the value of cc, we get the foci at (41,0)(\sqrt{41}, 0) and (−41,0)(-\sqrt{41}, 0).

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