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If a tangent to ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction blank+ fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction= 1 makes an angle alpha with x- axis, then square of length of intercept of tangent cut between axes is-

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  1. a to the power of 2 end exponent t a n to the power of 2 end exponent invisible function application alpha plus b to the power of 2 end exponent c o t to the power of 2 end exponent invisible function application alpha    
  2. a to the power of 2 end exponent s e c to the power of 2 end exponent invisible function application alpha plus b to the power of 2 end exponent c o s e c to the power of 2 end exponent invisible function application alpha    
  3. a to the power of 2 end exponent c o s e c to the power of 2 end exponent invisible function application alpha plus b to the power of 2 end exponent s e c to the power of 2 end exponent invisible function application alpha    
  4. None of these    

    Answer:The correct answer is: a to the power of 2 end exponent s e c to the power of 2 end exponent invisible function application alpha plus b to the power of 2 end exponent c o s e c to the power of 2 end exponent invisible function application alphaLet tangent is fraction numerator x cos invisible function application theta over denominator a end fraction+fraction numerator y sin invisible function application theta over denominator b end fraction= 1
    Slope is tan alpha = –fraction numerator b over denominator a end fraction c o t invisible function application theta S… (1)
    therefore square of length of intercept is
    = left parenthesis a s e c invisible function application theta right parenthesis to the power of 2 end exponent plus left parenthesis b c o s e c invisible function application theta right parenthesis to the power of 2 end exponent
    therefore Length is L equals a to the power of 2 end exponent plus b to the power of 2 end exponent plus a to the power of 2 end exponent t a n to the power of 2 end exponent invisible function application theta plus b to the power of 2 end exponent c o t to the power of 2 end exponent invisible function application theta … (2)
    Now use value of tan  from (1)

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