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What are the foci of the hyperbola represented by the equation (y^(2))/(41)-(x^(2))/(25)=1? Choose 1 answer: (A) (4,0) and (-4,0) (B) (sqrt66,0) and (-sqrt66,0) (c) (0,sqrt66) and (0,-sqrt66) (D) (0,4) and (0,-4)

What are the foci of the hyperbola represented by the equation y241−x225=1? \frac{y^{2}}{41}-\frac{x^{2}}{25}=1 ? \newlineChoose 11 answer:\newline(A) (4,0) (4,0) and (−4,0) (-4,0) \newline(B) (66,0) (\sqrt{66}, 0) and (−66,0) (-\sqrt{66}, 0) \newline(C) (0,66) (0, \sqrt{66}) and (0,−66) (0,-\sqrt{66}) \newline(D) (0,4) (0,4) and (0,−4) (0,-4)

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Q. What are the foci of the hyperbola represented by the equation y241−x225=1? \frac{y^{2}}{41}-\frac{x^{2}}{25}=1 ? \newlineChoose 11 answer:\newline(A) (4,0) (4,0) and (−4,0) (-4,0) \newline(B) (66,0) (\sqrt{66}, 0) and (−66,0) (-\sqrt{66}, 0) \newline(C) (0,66) (0, \sqrt{66}) and (0,−66) (0,-\sqrt{66}) \newline(D) (0,4) (0,4) and (0,−4) (0,-4)
  1. Hyperbola Equation Form: The given equation is in the form of a hyperbola with the equation y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1, where a2a^2 is under the y2y^2 term and b2b^2 is under the x2x^2 term. This indicates that the hyperbola opens up and down along the y-axis.
  2. Finding the Foci: To find the foci of the hyperbola, we need to use the formula c2=a2+b2c^2 = a^2 + b^2, where cc is the distance from the center to each focus.
  3. Calculating c^22: From the given equation, we can see that a2=41a^2 = 41 and b2=25b^2 = 25. Now we will calculate c2c^2.
  4. Substituting Values: Substitute the values of a2a^2 and b2b^2 into the formula to find c2c^2: c2=41+25c^2 = 41 + 25.
  5. Calculating c: Calculate c2c^2: c2=66c^2 = 66.
  6. Locating the Foci: Now, find the value of cc by taking the square root of c2c^2: c=66c = \sqrt{66}.
  7. Finding the Coordinates: Since the hyperbola opens up and down, the foci are located at (0,±c)(0, \pm c) along the y-axis.
  8. Finding the Coordinates: Since the hyperbola opens up and down, the foci are located at (0,±c)(0, \pm c) along the y-axis.Substitute the value of cc to find the coordinates of the foci: (0,66)(0, \sqrt{66}) and (0,−66)(0, -\sqrt{66}).

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