Maths-
General
Easy

Question

In an ellipse 9 x to the power of 2 end exponent plus 5 y to the power of 2 end exponent equals 45, the distance between the foci is

  1. 4 square root of 5    
  2. fraction numerator 49 over denominator 4 end fraction x to the power of 2 end exponent minus fraction numerator 51 over denominator 196 end fraction y to the power of 2 end exponent equals 1    
  3. 3    
  4. 4    

The correct answer is: 4


    Given equation may be written as fraction numerator x to the power of 2 end exponent over denominator 5 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 9 end fraction equals 1. Comparing the given equation with standard equation, we get a to the power of 2 end exponent equals 5 and b to the power of 2 end exponent equals 9. We also know that in an ellipse (where b to the power of 2 end exponent greater than a to the power of 2 end exponent right parenthesis a to the power of 2 end exponent equals b to the power of 2 end exponent left parenthesis 1 minus e to the power of 2 end exponent right parenthesis lambda or e to the power of 2 end exponent equals fraction numerator b to the power of 2 end exponent minus a to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals fraction numerator 9 minus 5 over denominator 9 end fraction equals fraction numerator 4 over denominator 9 end fraction or e equals fraction numerator 2 over denominator 3 end fraction.
    Therefore distance between foci equals 2 b e equals 2 cross times 3 cross times fraction numerator 2 over denominator 3 end fraction equals 4.

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