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Easy

Question

The equation of the hyperbola whose foci are (6, 4) and (–4, 4) and eccentricity 2 is given by

  1. 12 x to the power of 2 end exponent minus 4 y to the power of 2 end exponent minus 24 x plus 32 y minus 127 equals 0    
  2. 12 x to the power of 2 end exponent plus 4 y to the power of 2 end exponent plus 24 x minus 32 y minus 127 equals 0    
  3. 12 x to the power of 2 end exponent minus 4 y to the power of 2 end exponent minus 24 x minus 32 y plus 127 equals 0    
  4. 12 x to the power of 2 end exponent minus 4 y to the power of 2 end exponent plus 24 x plus 32 y plus 127 equals 0    

The correct answer is: 12 x to the power of 2 end exponent minus 4 y to the power of 2 end exponent minus 24 x plus 32 y minus 127 equals 0


    Foci are (6,4) and (–4,4), e equals 2 and centre is open parentheses fraction numerator 6 minus 4 over denominator 2 end fraction comma 4 close parentheses equals left parenthesis 14 right parenthesis
    Þ6 equals 1 plus a e Þ a e equals 5Þ a equals fraction numerator 5 over denominator 2 end fractionand b equals fraction numerator 5 over denominator 2 end fraction left parenthesis square root of 3 right parenthesis
    Hence the required equation is fraction numerator left parenthesis x minus 1 right parenthesis to the power of 2 end exponent over denominator left parenthesis 25 divided by 4 right parenthesis end fraction minus fraction numerator left parenthesis y minus 4 right parenthesis to the power of 2 end exponent over denominator left parenthesis 75 divided by 4 right parenthesis end fraction equals 1
    or 12 x to the power of 2 end exponent minus 4 y to the power of 2 end exponent minus 24 x plus 32 y minus 127 equals 0.

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