Why does ratio of $2$ linear expression gives you a rectangular hyperbola

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We were studying Calculus and some methods on how to find domain and range, when my teacher suddenly said, "FYI, the ratio of $2$ linear expression gives you a rectangular hyperbola



Can someone tell why does that give a "rectangular hyperbola only?"




calculus algebra-precalculus conic-sections




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    You can re-arrange the equation to $(x-a)(y-b)=c$,
    – Lord Shark the Unknown
    Jul 22 at 17:53










  • In order to answer such a question we should know the following: What is your definition of a hyperbola?
    – Christian Blatter
    Jul 22 at 18:35














up vote
0
down vote

favorite












We were studying Calculus and some methods on how to find domain and range, when my teacher suddenly said, "FYI, the ratio of $2$ linear expression gives you a rectangular hyperbola



Can someone tell why does that give a "rectangular hyperbola only?"




calculus algebra-precalculus conic-sections




share|cite|improve this question















  • 2




    You can re-arrange the equation to $(x-a)(y-b)=c$,
    – Lord Shark the Unknown
    Jul 22 at 17:53










  • In order to answer such a question we should know the following: What is your definition of a hyperbola?
    – Christian Blatter
    Jul 22 at 18:35












up vote
0
down vote

favorite









up vote
0
down vote

favorite











We were studying Calculus and some methods on how to find domain and range, when my teacher suddenly said, "FYI, the ratio of $2$ linear expression gives you a rectangular hyperbola



Can someone tell why does that give a "rectangular hyperbola only?"




calculus algebra-precalculus conic-sections




share|cite|improve this question











We were studying Calculus and some methods on how to find domain and range, when my teacher suddenly said, "FYI, the ratio of $2$ linear expression gives you a rectangular hyperbola



Can someone tell why does that give a "rectangular hyperbola only?"





calculus algebra-precalculus conic-sections





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 22 at 17:41









William

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753214







  • 2




    You can re-arrange the equation to $(x-a)(y-b)=c$,
    – Lord Shark the Unknown
    Jul 22 at 17:53










  • In order to answer such a question we should know the following: What is your definition of a hyperbola?
    – Christian Blatter
    Jul 22 at 18:35












  • 2




    You can re-arrange the equation to $(x-a)(y-b)=c$,
    – Lord Shark the Unknown
    Jul 22 at 17:53










  • In order to answer such a question we should know the following: What is your definition of a hyperbola?
    – Christian Blatter
    Jul 22 at 18:35







2




2




You can re-arrange the equation to $(x-a)(y-b)=c$,
– Lord Shark the Unknown
Jul 22 at 17:53




You can re-arrange the equation to $(x-a)(y-b)=c$,
– Lord Shark the Unknown
Jul 22 at 17:53












In order to answer such a question we should know the following: What is your definition of a hyperbola?
– Christian Blatter
Jul 22 at 18:35




In order to answer such a question we should know the following: What is your definition of a hyperbola?
– Christian Blatter
Jul 22 at 18:35










1 Answer
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$$ dfracax + bcx+d = dfracfracac left(cx + d right) - fracadc + bcx+d = dfracac + dfracbc-adc cdot dfrac1cx+d$$



Setting $dfracac = K $ and $dfracbc-adc= A $, we get,



$K + dfracAcx+d $ which is basically of the form, $ y = dfracAx$, this is the equation of rectangular hyperbola whose asymptotes are co-ordinate axes.



Check more about that on, https://en.m.wikipedia.org/wiki/Hyperbola#Hyperbola_with_equation_y=A/x






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    $$ dfracax + bcx+d = dfracfracac left(cx + d right) - fracadc + bcx+d = dfracac + dfracbc-adc cdot dfrac1cx+d$$



    Setting $dfracac = K $ and $dfracbc-adc= A $, we get,



    $K + dfracAcx+d $ which is basically of the form, $ y = dfracAx$, this is the equation of rectangular hyperbola whose asymptotes are co-ordinate axes.



    Check more about that on, https://en.m.wikipedia.org/wiki/Hyperbola#Hyperbola_with_equation_y=A/x






    share|cite|improve this answer

























      up vote
      3
      down vote













      $$ dfracax + bcx+d = dfracfracac left(cx + d right) - fracadc + bcx+d = dfracac + dfracbc-adc cdot dfrac1cx+d$$



      Setting $dfracac = K $ and $dfracbc-adc= A $, we get,



      $K + dfracAcx+d $ which is basically of the form, $ y = dfracAx$, this is the equation of rectangular hyperbola whose asymptotes are co-ordinate axes.



      Check more about that on, https://en.m.wikipedia.org/wiki/Hyperbola#Hyperbola_with_equation_y=A/x






      share|cite|improve this answer























        up vote
        3
        down vote










        up vote
        3
        down vote









        $$ dfracax + bcx+d = dfracfracac left(cx + d right) - fracadc + bcx+d = dfracac + dfracbc-adc cdot dfrac1cx+d$$



        Setting $dfracac = K $ and $dfracbc-adc= A $, we get,



        $K + dfracAcx+d $ which is basically of the form, $ y = dfracAx$, this is the equation of rectangular hyperbola whose asymptotes are co-ordinate axes.



        Check more about that on, https://en.m.wikipedia.org/wiki/Hyperbola#Hyperbola_with_equation_y=A/x






        share|cite|improve this answer













        $$ dfracax + bcx+d = dfracfracac left(cx + d right) - fracadc + bcx+d = dfracac + dfracbc-adc cdot dfrac1cx+d$$



        Setting $dfracac = K $ and $dfracbc-adc= A $, we get,



        $K + dfracAcx+d $ which is basically of the form, $ y = dfracAx$, this is the equation of rectangular hyperbola whose asymptotes are co-ordinate axes.



        Check more about that on, https://en.m.wikipedia.org/wiki/Hyperbola#Hyperbola_with_equation_y=A/x







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 22 at 17:47









        Ibrahim

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