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Sorry for the bad title , please edit it to something better if you can.

I need a procedure to convert the general equation of ellipse - $$Ax^2 + By^2 + 2hxy + 2gx + 2fy + c = 0$$ into $$\frac{\left \{ \frac{ux+vy+q}{\sqrt{u^{2}+v^{2}}} \right \}}{a^{2}}^{2} + \frac{\left \{ \frac{vx-uy+w}{\sqrt{u^{2}+v^{2}}} \right \}}{b^{2}}^{2}=1$$ which is like $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$

(Without rotating or translating . I know about rotating and translating to convert it into the above standard form (last equation) , that's not what I want).

A Googler
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  • What is wrong with the "rotating and translating" method that causes you not to want to use it? – David K Sep 24 '15 at 19:16
  • @DavidK I don't want to change the coordinate system . $ux +vy+q$ and $vx-uy+w$ are axes of ellipse so they're like $ x$ and $y$ of standard ellipse. – A Googler Sep 24 '15 at 19:18
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    OK, so when you said "convert it into the above standard form" you meant the last equation you wrote, not the one with $u$ and $v$ in it. You only know how to rotate and translate to get the last equation. Is that right? – David K Sep 24 '15 at 19:20
  • @DavidK Yes , that's right. – A Googler Sep 24 '15 at 19:20
  • Have you tried expanding the second expression and equating coefficients? – robjohn Sep 24 '15 at 19:27
  • If you know the amount of rotation $\theta$ and translation $(\Delta x, \Delta y)$ that would get the last equation, you can easily derive the equation with $u$ and $v$ from $\theta$, $\Delta x$, and $\Delta y$. If you don't know $\theta$, $\Delta x$, and $\Delta y$ already then it takes more work (at least that's what I recall from the last time I did this), but robjohn gave a good hint. – David K Sep 24 '15 at 19:29
  • An ellipse that is NOT in "standard form" in a given coordinate system is not in "standard position" (center at (0, 0) and axes of symmetry along the coordinate axes) can only be in standard form in a new coordinate system in which it [b]is[/b] in standard form. So you [b]have[/b] to change to a new coordinate system. You can then, if you like, write the ellipse equation in terms of the original coordinate system. – user247327 Sep 24 '15 at 19:31
  • @robjohn Hmm that seems hard to do by hand. Maybe if someone has any software that can do this , that would help. – A Googler Sep 24 '15 at 19:33
  • @AGoogler: that is how I did something similar many years ago (long before there were personal computers). – robjohn Sep 24 '15 at 19:40
  • See the MathWorld page on "Ellipse"; starting with the general equation $ax^2+2bxy+cy^2+2dx+2fy+g=0$ (which at least at this moment is equation 15 on that page), it finds the center, semiaxis lengths, and angle of rotation. – David K Sep 24 '15 at 20:35
  • You don't rotate or translate the ellipse.You describe the ellipse in terms of co-ordinates with respect to a different pair of axes. – DanielWainfleet Sep 24 '15 at 22:45

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