text: # find the equationof the hyperbola satifying the give conditions: Foci (4, 0), the latus rectum is of length 12.
  
  Latika Leekha
  9 years ago
  Let the equation of hyperbola is x2/a2 – y2/b2 = 1.
  where b2 = a2 (e2 -1).
  For hyperbola, the eccentricity e > 1.
  The focus is given as (4, 0). Hence, we have ae = 4.
  This gives a = 4/e. … (1)
  Latus Rectum = 2b2/ a = 12.
  Hence, b2 = 6a. …... (2)
  since, b2 = a2 (e2 -1)
  Using (1) and (2) in the above equation, we get
  6a = a2(e2 -1)
  4/e = (4/e)2 (e2 -1)
  Simplifying, we get
  4e2 – 6e – 4 = 0
  (2e +1)(e – 2) = 0.
  This gives e = -1/2 or 2.
  Hence, we have e = 2. ( since e > 1 in case of hyperbola).
  hence, we have a = 4/e = 4/2 = 2.
  Substituting this value we get b2 = 6a = 6.2 = 12
  Hence, the eqauation of hyperbola is x2/a2 – y2/b2 = 1
  x2/4 – y2/12 = 1.
  Simplifying we get, the required hyperbola as
  3x2– y2 = 12.
  Thanks & Regards
  Latika Leekha
explanation: The document contains a math question about finding the equation of a hyperbola given its foci and the length of the latus rectum. The document also includes a detailed step-by-step solution to the problem.
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distilabel_metadata.role: system
distilabel_metadata.content: # find the equationof the hyperbola satifying the give conditions: Foci (4, 0), the latus rectum is of length 12.
  
  Latika Leekha
  9 years ago
  Let the equation of hyperbola is x2/a2 – y2/b2 = 1.
  where b2 = a2 (e2 -1).
  For hyperbola, the eccentricity e > 1.
  The focus is given as (4, 0). Hence, we have ae = 4.
  This gives a = 4/e. … (1)
  Latus Rectum = 2b2/ a = 12.
  Hence, b2 = 6a. …... (2)
  since, b2 = a2 (e2 -1)
  Using (1) and (2) in the above equation, we get
  6a = a2(e2 -1)
  4/e = (4/e)2 (e2 -1)
  Simplifying, we get
  4e2 – 6e – 4 = 0
  (2e +1)(e – 2) = 0.
  This gives e = -1/2 or 2.
  Hence, we have e = 2. ( since e > 1 in case of hyperbola).
  hence, we have a = 4/e = 4/2 = 2.
  Substituting this value we get b2 = 6a = 6.2 = 12
  Hence, the eqauation of hyperbola is x2/a2 – y2/b2 = 1
  x2/4 – y2/12 = 1.
  Simplifying we get, the required hyperbola as
  3x2– y2 = 12.
  Thanks & Regards
  Latika Leekha
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question about finding the equation of a hyperbola given its foci and the length of the latus rectum. The document also includes a detailed step-by-step solution to the problem.
  
  Label: question-and-answer