Symmetry problem

  • #1
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Homework Statement


Find the point in respect to these ellipses are symmetrical
x2 + 3y2 - 2x -2 = 0
x2 + 3y2 +6x + 12y +18 = 0


Homework Equations


x = 2a - x'
y = 2b - y'

The Attempt at a Solution


I have applied the equations of symmetry to the first equation then I've equaled the result to the second ellipse
4a2 + 12b2 - 4ax - 12bx - 4a -4x -20 -12y = 0

I don't know how to solve this for a and b or if I have mistaken something...
thanks
 

Answers and Replies

  • #2
SammyS
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Homework Statement


Find the point in respect to these ellipses are symmetrical
x2 + 3y2 - 2x -2 = 0
x2 + 3y2 +6x + 12y +18 = 0

Homework Equations


x = 2a - x'
y = 2b - y'

The Attempt at a Solution


I have applied the equations of symmetry to the first equation then I've equaled the result to the second ellipse
4a2 + 12b2 - 4ax - 12bx - 4a -4x -20 -12y = 0

I don't know how to solve this for a and b or if I have mistaken something...
thanks
Precisely, what is it you are trying to do here?

What is the full statement of the problem?
 
  • #3
SammyS
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Write the equation of each ellipse in standard form.

##\displaystyle \frac{(x-h)^2}{A^2}+\frac{(y-k)^2}{B^2}=1##
 
  • #4
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I'm trying to find the symmetry point of that two ellipses, it should be (-1;-1) to find it I have to solve that for a and b but how ?
 
  • #5
SammyS
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I'm trying to find the symmetry point of that two ellipses, it should be (-1;-1) to find it I have to solve that for a and b but how ?
Yes. About 5 minutes before I saw your post, I figured out what you're trying to do.

Find the point of symmetry such that if you reflect the points on either ellipse through this point of symmetry, the points will be on the other ellipse.

You have a typo in the equation:
4a2 + 12b2 - 4ax - 12bx - 4a -4x -20 -12y = 0​

That should be
4a2 + 12b2 - 4ax - 12by - 4a -4x -20 -12y = 0​

The only way to make this be zero is for the coefficients of x and of y to be zero.

[STRIKE]It looks as if there is another error. The constant term must also be zero.[/STRIKE]

Added in Edit: The constant term will also be zero.
 
Last edited:
  • #6
tiny-tim
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hi scientifico! :smile:
x2 + 3y2 - 2x -2 = 0
x2 + 3y2 +6x + 12y +18 = 0
I'm trying to find the symmetry point of that two ellipses, it should be (-1;-1)

the two ellipses are the same shape

if you know where their centres are, isn't it obvious where the symmetry point is? :confused:
 
  • #7
haruspex
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I have applied the equations of symmetry to the first equation then I've equaled the result to the second ellipse
4a2 + 12b2 - 4ax - 12bx - 4a -4x -20 -12y = 0
I get a different equation. Pls post your working.
Once you have the right equation, you are looking for values of a and b that make the equation true for all x and y. (That can't work for the equation you got because the y coefficient is a nonzero constant.) Can you see how to use that?
 

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