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 $b=1$.

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?

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

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

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

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

• Thank you! How do I know in which direction to rotate? May 22 '20 at 15:24
• @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. May 22 '20 at 15:31
• So we rotate to the angle equal to the angle between major axis and the $x$-axis, right? May 22 '20 at 15:33
• @VIVID that is correct May 22 '20 at 15:34
• 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... May 22 '20 at 15:40

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!

• 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.
– amd
May 22 '20 at 18:30

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.