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Find the equation of the ellipse whose symmetry axes are given by $x+y-2=0$ and $y-x-1=0$. Also semi width $a=2$ and semi height $b=1$.

As the center can be found at the intersection of the symmetry axes, I found that the center is at $C\big(\frac{1}{2},\frac{3}{2}\big)$. As the center is symmetry center for ellipse, symmetrical points on the first and the second symmetry axes do also belong to the ellipse. Those points are $2$ and $1$ units far from the center. Hence, they are of the form $$A_1\big(\frac{1}{2}-\sqrt{2},\frac{3}{2}+\sqrt{2}\big), \ \ A_2\big(\frac{1}{2}+\sqrt{2},\frac{3}{2}-\sqrt{2}\big), \ \ B_1\big(\frac{1}{2}-\frac{1}{\sqrt{2}},\frac{3}{2}-\frac{1}{\sqrt{2}}\big), \ \ B_2\big(\frac{1}{2}+\frac{1}{\sqrt{2}},\frac{3}{2}+\frac{1}{\sqrt{2}}\big)$$ Now I have 5 points that lie on the ellipse of a general form: $$Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0$$ And by plugging them into the equation I may probably find the coefficients. However, this seems to be a lot of irrational computations :(

Could anyone suggest me a wiser way to solve this problem?

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    $\begingroup$ $(\frac{(x-\frac12)+(y-\frac32)}{\sqrt{2}})^2/2^2+(\frac{(y-\frac32)-(x-\frac12)}{\sqrt{2}})^2=1$ or $(\frac{(x-\frac12)+(y-\frac32)}{\sqrt{2}})^2+(\frac{(y-\frac32)-(x-\frac12)}{\sqrt{2}})^2/2^2=1$ $\endgroup$ May 22 '20 at 13:23
  • $\begingroup$ @Jan-MagnusØkland How did you find this? $\endgroup$
    – VIVID
    May 22 '20 at 13:24
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    $\begingroup$ I wrote the lines in a form that conserves the distances, then it's just choosing which semi-axis has length what. $\endgroup$ May 22 '20 at 13:26
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Start with the canonical equation below

$$\frac{x^2}2+ \frac{y^2}1=1$$

First, rotate 45 degrees, i.e. $x\to \frac1{\sqrt2}(x+y)$, $y\to \frac1{\sqrt2}(-x+y)$

$$\frac{(x+y)^2}4+ \frac{(x-y)^2}2=1$$

Then, shift the center $x\to x-\frac12$ and $y\to y-\frac32$ to obtain

$$\frac{(x+y-2)^2}4+ \frac{(x-y+1)^2}2=1$$

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  • $\begingroup$ Thank you! How do I know in which direction to rotate? $\endgroup$
    – VIVID
    May 22 '20 at 15:24
  • $\begingroup$ @VIVID - There are two possibilities. I just assumed that the major axis is at 45-degrees with the x-axis, while the other is at 135-degrees. $\endgroup$
    – Quanto
    May 22 '20 at 15:31
  • $\begingroup$ So we rotate to the angle equal to the angle between major axis and the $x$-axis, right? $\endgroup$
    – VIVID
    May 22 '20 at 15:33
  • $\begingroup$ @VIVID that is correct $\endgroup$
    – Quanto
    May 22 '20 at 15:34
  • $\begingroup$ Then if you take this angle equal to 45-degrees, your rotation matrix would be of the form $$\begin{bmatrix} \cos 45^{\circ} & -\sin 45^{\circ} \\ \sin 45^{\circ} & \cos 45^{\circ} \end{bmatrix}$$ So that $x \rightarrow \frac{1}{\sqrt{2}}(x-y)$ and $y \rightarrow \frac{1}{\sqrt{2}}(x+y)$. But you got something little bit different... $\endgroup$
    – VIVID
    May 22 '20 at 15:40
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I think this would be an easier way:First I want to talk about simpler ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ we know that major axis of this ellipse is $y=0$ and minor axis of the ellipse is $x=0$.Now i am trying to see the equation in terms of these two axes.If we observe $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ clearly we can get that it's more like $$\frac{(\text{distance of general point}(x,y) \text{ from minor axis})^2}{(\text{lenght of semi major axis})^2}+\frac{(\text{distance of general point}(x,y) \text{ from major axis})^2}{(\text{lenght of semi minor axis})^2}=1$$ and i think applying this techinque to this question would be of a great help.Hope that helps!

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  • $\begingroup$ I think this idea might be a bit clearer if you pointed out that the $x$-coordinate of a point is its (signed) distance from the $y$-axis and vice-versa. This answer is in much the same vein. $\endgroup$
    – amd
    May 22 '20 at 18:30
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HINTS:

The slopes of given lines are, using slope-intercept form are $ =\pm1$, a rotation by $ + 45^{\circ}.$

By distance formula for end points of major/minor axes you found is calculated as

$$ d_{A_1A_2}^2= ( 2 \sqrt2)^2+ (2 \sqrt2)^2 = 16 \tag1 $$

so the major axis length is ( square root and half) $ \;a=2$

Similarly $ b=1 $

Quite beneficial to adopt parametric form

$$ ( x,y)= ( a \cos t ,b \sin t) \tag 2 ) $$

First rotation by $ \pi/4$ standard rotation relation:

$$ ( x_1,y_1) = [(x-y)/\sqrt2/,(x+y)/\sqrt2 ]\tag3 $$

Next displacement of ellipse center to $ C (\dfrac12, \dfrac32) =(h,k) $ say

$$x_2=( x_1+h); \; y_2=(y_1+k) \tag4$$

If you want to convert to $(x,y) $ form by trig elimination of $t$, that is up to you.

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