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Hyperbola question

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  1. Jun 4, 2015 #1
    Back in 10th degree, I have learned that in Clapeyron-Mendeleev coordinates ( eq: p-V) , an Isotherm transformation of an ideal gas ( with constant mass throughout the transformation ) is represented with an arc of an hyperbola. Now, I have learned that hyperbola equation is : x2 / a2 - y2/b2 = 1 ( or written in the other way, with y2 as first term ) . This equation , plotted, result in a different type of graphic as I learned on T=constant transformation ! My question is why I used to draw the curbe line graph in p-V coordinates of an equation like y=1/x ? ( as pV= constant ) , saying that it is a hyperbola? What has got to do with an arc of hyperbola? Thank you !
     
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  3. Jun 4, 2015 #2

    DEvens

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    If you have the curve given by ## x^2 - y^2 = 1## and you rotate it 45 degrees around the origin, what do you get?
     
  4. Jun 4, 2015 #3

    robphy

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    As a variant of DEvens's post, suppose you define u=x+y and v=x-y. What is the curve in terms of u and v?
     
  5. Jun 5, 2015 #4
    Thank you for your responses !
    @DEvens :yes, I realised that x^2−y^2=1 shifted by π/4 radians results in my desired part of graph, but... something is not clear in my mind. How can I assume that I can rotate the graph and still get something mathematically valid? There's something not clear in my mind...
    @robphy : in terms of u and v, u*v = x^2-y^2 =constant =1 , as in a normal isotherm. But from the (x,y) coordinates , I can define a new system of coordinates, given by (x+y, x-y ) ? Just like when I shift by π/4 radians the normal hyperbola graph?
     
  6. Jun 5, 2015 #5

    robphy

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    A hyperbola is shape, not a function.
    Focus on the curve, and forget about the axes.

    Drawn on a piece of paper, that curve is a hyperbola... no matter how you slide, reflect, or rotate the piece of paper.
     
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