How to find the equations of the axis of the ellipse

[pbialos]

I was hoping you could give me a hint on how to find the equations of the axis of the ellipse of equation $$5x^2-6xy+5y^2-4x-4y-4=0$$. I think this is supposed to be an exercise about lagrange multipliers or something related to the gradient, but i really don't know. I am clueless, i don't know any property about ellipses in general.

Many Thanks, Paul.

 

[TD]

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 $m_1 = - \frac{1}{{m_2 }}$, where m is a direction.

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

In this case, you get $m = 1\,\,\, \vee \,\,m = - 1$.

Now, the equations of the axes are then:

$$\begin{array}{l}<br /> F_x ^\prime \left( {x,y} \right) + m_1 \cdot F_y ^\prime \left( {x,y} \right) = 0 \\ <br /> F_x ^\prime \left( {x,y} \right) + m_2 \cdot F_y ^\prime \left( {x,y} \right) = 0 \\ <br /> \end{array}$$

Here, $F_x ^\prime \left( {x,y} \right)$ 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?

 

[thepatientmental]

Hi Paul,

To find the equations of the axes of an ellipse, you can start by rewriting the given equation in standard form, which is (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k) is the center of the ellipse and a and b are the lengths of the semi-major and semi-minor axes, respectively.

To do this, you can use the method of completing the square to group the terms with x and y together. Once you have the equation in standard form, you can easily identify the center and lengths of the axes.

In this case, the equation can be rewritten as (5x^2-4x) + (-6xy+5y^2-4y) = 4. Then, completing the square for the terms with x and y, we get (5(x^2-4/5x)) + (-6(x^2-4/5xy)) + (5(y^2-4/5y)) = 4. Simplifying and factoring, we get (5(x-2/5)^2) + (-6(x-2/5)(y-2/3)) + (5(y-2/3)^2) = 4.

Now, comparing this to the standard form equation, we can see that the center is at (2/5, 2/3) and the lengths of the axes are (5/√4) and (5/√5). Therefore, the equations of the axes are x-2/5 = 0 and y-2/3 = 0.

I hope this helps! Remember to always look for patterns and use the properties of ellipses when solving these types of problems.