How to find the equations of the axis of the ellipse

  • Thread starter pbialos
  • Start date
  • #1
pbialos
I was hoping you could give me a hint on how to find the equations of the axis of the ellipse of equation [tex]5x^2-6xy+5y^2-4x-4y-4=0[/tex]. I think this is supposed to be an exercise about lagrange multipliers or something related to the gradient, but i really dont know. I am clueless, i dont know any property about ellipses in general.

Many Thanks, Paul.
 

Answers and Replies

  • #2
TD
Homework Helper
1,022
0
Unfortunately, I don't know the English terminology that well regarding analytic geometry.

There is a thing which we call "middle line", which is a polar (polar line) of a point at infinity with respect to the conic (here the ellipse).

Axes are a special case of these lines, two of those whose directions are perpendicular, so where [itex]m_1 = - \frac{1}{{m_2 }}[/itex], where m is a direction.

For a general conic [itex]ax^2 + 2b''xy + a'y + 2by + 2b'x + a'' = 0[/itex] those direction are the solutions of [itex]b''m^2 + \left( {a - a'} \right)m - b'' = 0[/itex].

In this case, you get [itex]m = 1\,\,\, \vee \,\,m = - 1[/itex].

Now, the equations of the axes are then:

[tex]\begin{array}{l}
F_x ^\prime \left( {x,y} \right) + m_1 \cdot F_y ^\prime \left( {x,y} \right) = 0 \\
F_x ^\prime \left( {x,y} \right) + m_2 \cdot F_y ^\prime \left( {x,y} \right) = 0 \\
\end{array}[/tex]

Here, [itex]F_x ^\prime \left( {x,y} \right)[/itex] mean the partial derivative of the function to the variable x (same for y).

I tried it and it seems to be working, can you get the equations now?
 

Related Threads on How to find the equations of the axis of the ellipse

Replies
19
Views
14K
Replies
8
Views
24K
Replies
4
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
5
Views
9K
  • Last Post
Replies
8
Views
3K
Replies
0
Views
1K
  • Last Post
Replies
1
Views
5K
Replies
5
Views
6K
Top