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 $b=1$.

Question

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?

Answer 1

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

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

Answer 2

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!

Answer 3

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.