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Easy

Question

The locus of the foot of perpendicular drawn from the centre text  of the ellipse  end text x to the power of 2 end exponent plus 3 y to the power of 2 end exponent equals 6 text  on any tangent to it is end text

  1. open parentheses x to the power of 2 end exponent plus y to the power of 2 end exponent close parentheses to the power of 2 end exponent equals 6 x to the power of 2 end exponent plus 2 y to the power of 2 end exponent    
  2. open parentheses x to the power of 2 end exponent plus y to the power of 2 end exponent close parentheses to the power of 2 end exponent equals 6 x to the power of 2 end exponent minus 2 y to the power of 2 end exponent    
  3. open parentheses x to the power of 2 end exponent minus y to the power of 2 end exponent close parentheses to the power of 2 end exponent equals 6 x to the power of 2 end exponent plus 2 y to the power of 2 end exponent    
  4. open parentheses x to the power of 2 end exponent minus y to the power of 2 end exponent close parentheses to the power of 2 end exponent equals 6 x to the power of 2 end exponent minus 2 y to the power of 2 end exponent    

The correct answer is: open parentheses x to the power of 2 end exponent plus y to the power of 2 end exponent close parentheses to the power of 2 end exponent equals 6 x to the power of 2 end exponent plus 2 y to the power of 2 end exponent


    See Fig. 14.20. Ellipse is fraction numerator x to the power of 2 end exponent over denominator left parenthesis square root of 6 right parenthesis to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator left parenthesis square root of 2 right parenthesis to the power of 2 end exponent end fraction equals 1
    text Equation of tangent at  end text P text  is  end text fraction numerator x c o s invisible function application theta over denominator square root of 6 end fraction plus fraction numerator y s i n invisible function application theta over denominator square root of 2 end fraction equals 1

    text Equation of tangent at  end text left parenthesis alpha comma beta right parenthesis text  is end text
    y minus beta equals negative fraction numerator alpha over denominator beta end fraction left parenthesis x minus alpha right parenthesis text  or  end text beta y minus beta to the power of 2 end exponent equals negative alpha x plus alpha to the power of 2 end exponent
    table row cell text end text text O end text text r end text text end text end cell row cell alpha x plus beta y equals alpha to the power of 2 end exponent plus beta to the power of 2 end exponent end cell end table
    text Comparing Eq. (1) and Eq. (2), we get end text
    fraction numerator cos invisible function application theta over denominator square root of 6 alpha end fraction equals fraction numerator sin invisible function application theta over denominator square root of 2 beta end fraction equals fraction numerator 1 over denominator alpha to the power of 2 end exponent plus beta to the power of 2 end exponent end fraction
    table row cell text end text text O end text text r end text text end text end cell row cell square root of fraction numerator c o s to the power of 2 end exponent invisible function application theta plus s i n to the power of 2 end exponent invisible function application theta over denominator 6 alpha to the power of 2 end exponent plus 2 beta to the power of 2 end exponent end fraction end root rightwards double arrow fraction numerator 1 over denominator square root of 6 alpha to the power of 2 end exponent plus 2 beta to the power of 2 end exponent end root end fraction equals fraction numerator 1 over denominator alpha to the power of 2 end exponent plus beta to the power of 2 end exponent end fraction end cell end table
    rightwards double arrow 6 alpha to the power of 2 end exponent plus 2 beta to the power of 2 end exponent equals open parentheses alpha to the power of 2 end exponent plus beta to the power of 2 end exponent close parentheses to the power of 2 end exponent
    text Therefore, locus is  end text open parentheses x to the power of 2 end exponent plus y to the power of 2 end exponent close parentheses to the power of 2 end exponent equals 6 x to the power of 2 end exponent plus 2 y to the power of 2 end exponent text  . end text

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