Maths-
General
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Question

If a hyperbola passing through the origin has 3 x minus 4 y minus 1 equals 0 and 4 x minus 3 y minus 6 equals 0 as its asymptotes, then the equation of its tranvsverse and conjugate axes are

  1. x minus y minus 5 equals 0 and x plus y plus 1 equals 0    
  2. x minus y equals 0 and x plus y plus 5 equals 0    
  3. x plus y minus 5 equals 0 and x minus y minus 1 equals 0    
  4. x plus y minus 1 equals 0 and x minus y minus 5 equals 0    

Hint:

An asymptote of a curve in analytical geometry is a line where the distance between the curve and the line tends to zero. Here we have given hyperbola passing through the origin has 3 x minus 4 y minus 1 equals 0 and 4 x minus 3 y minus 6 equals 0 as its asymptotes, we have to find the equation of its transverse and conjugate axes.

The correct answer is: x plus y minus 5 equals 0 and x minus y minus 1 equals 0


    a line or curve that serves as the boundary of another line or curve in mathematics. An example of an asymptotic curve is a descending curve that approaches but does not reach the horizontal axis, which is the asymptote of the curve.
    The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.
    So we have:
    fraction numerator 3 x minus 4 y minus 1 over denominator square root of 3 squared plus left parenthesis negative 4 right parenthesis squared end root end fraction equals plus space fraction numerator 4 x minus 3 y minus 6 over denominator square root of 4 squared plus left parenthesis negative 3 right parenthesis squared end root end fraction
fraction numerator 3 x minus 4 y minus 1 over denominator square root of 25 end fraction equals plus space fraction numerator 4 x minus 3 y minus 6 over denominator square root of 25 end fraction
3 x minus 4 y minus 1 equals 4 x minus 3 y minus 6
3 x minus 4 y minus 1 minus 4 x plus 3 y plus 6 equals 0
x plus y minus 5 equals 0
T h i s space i s space t h e space e q u a t i o n space o f space t r a n s v e r s e.
fraction numerator 3 x minus 4 y minus 1 over denominator square root of 3 squared plus left parenthesis negative 4 right parenthesis squared end root end fraction equals negative space fraction numerator 4 x minus 3 y minus 6 over denominator square root of 4 squared plus left parenthesis negative 3 right parenthesis squared end root end fraction
fraction numerator 3 x minus 4 y minus 1 over denominator square root of 25 end fraction equals negative space fraction numerator 4 x minus 3 y minus 6 over denominator square root of 25 end fraction
3 x minus 4 y minus 1 equals negative space 4 x plus 3 y plus space 6
3 x minus 4 y minus 1 plus space 4 x minus 3 y minus space 6 equals 0
7 x minus 7 y equals 7
x minus y minus 1 equals 0
T h i s space i s space e q u a t i o n space o f space c o n j u g a t e space a x i s.

    Here we used the concept of polar coordinate system and also the trigonometric ratios to find the solution. T h e space e q u a t i o n space o f space t r a n s v e r s e space i s space x plus y minus 5 equals 0. space T h e space e q u a t i o n space o f space c o n j u g a t e space a x i s space i s space x minus y minus 1 equals 0.

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