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What happens when a hyperbola is tilted and we have to find its standard equation?

For example, take the conversion in this image when the hyperbola converted from the first form to the next form. my guess for the conditions of the conversion were

  1. both the lines(in the bracket) had to be perpendicular to each other
  2. it had to be like a unit vector. but I don't get why the second equation had root3 instead of root2

Are my guesses correct? If so can you tell me the values of a nd b for the above equation?

aaravm
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  • It looks like they are going for a rotation of axes through 45 degrees, and tried to rewrite in the rotated variables. But if so it seems incorrect to me, as rotation through 45 degrees makes the rotated coordinates U=(x-y)/sqrt(2) and V=(x+y)/sqrt(2), in order to have the rotation preserve distances. I also don't see how they are using sqrt(3). Again, to clear things up here you should quote the entire set of manipulations the text is doing. And it should be put in the question not a link to a photo. – coffeemath Apr 21 '22 at 05:37
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    The manipulation in the image is not helpful if the intent is to find $a, b$.

    $$5y^2-26xy+5x^2=36$$

    $$(x-y)^2/2^2-(x+y)^2/3^2=1$$

    $$\frac{\frac{(x-y)^2}{2}}{2}-\frac{\frac{(x+y)^2}{2}}{3^2/2}=1$$

    Then you can do the orthonormal coordinate change

    $$x'=\frac{x+y}{\sqrt2}, y'=\frac{x-y}{\sqrt2}$$

    $$\frac{y'^2}{\sqrt2^2}-\frac{x'^2}{(3/\sqrt2)^2}=1$$

    $$y'^2/b^2-x'^2/a^2=1,,0<b=\sqrt2<a=3/\sqrt2,$$

    the conjugate hyperbola of the one usually described, having the same asymptotes, but on the other side.

     – Jan-Magnus Økland Apr 21 '22 at 06:45

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