- #1
Hugo S
- 3
- 0
This is a question I have been playing with this week out of curiosity but I keep coming up against brick walls and unenlightening results.
Given the equation of an ellipse, say $$ x^{2} - xy + y^2 = 2, $$ I would like to find the equation of the line which passes through the major axis.
I tried solving for x and y and applying a rotation matrix. Doing so, I found that
$$x' = \cos \theta (-\sqrt{8-3y^2} - y) + \sin \theta (-\sqrt{8 - 3x^2} - x) $$
$$y' = \sin \theta (\sqrt{8-3y^2} + y) + \cos \theta (-\sqrt{8 - 3x^2} - x). $$
Setting the second equation to equal zero and solving in terms of θ I ended up with the result: $$ \tan \theta = \frac{\sqrt{8-3x^2} + x}{\sqrt{8-3y^2 + y}}. $$
This result doesn't seem to help me as it is ultimately telling me something very trivial that I already know about the relationship between x and y coordinates and the tan function.
I know that my choice of coordinates might not be particularly strategic. Are there any other mistakes I'm making in how I'm approaching this?
Thank you!
Given the equation of an ellipse, say $$ x^{2} - xy + y^2 = 2, $$ I would like to find the equation of the line which passes through the major axis.
I tried solving for x and y and applying a rotation matrix. Doing so, I found that
$$x' = \cos \theta (-\sqrt{8-3y^2} - y) + \sin \theta (-\sqrt{8 - 3x^2} - x) $$
$$y' = \sin \theta (\sqrt{8-3y^2} + y) + \cos \theta (-\sqrt{8 - 3x^2} - x). $$
Setting the second equation to equal zero and solving in terms of θ I ended up with the result: $$ \tan \theta = \frac{\sqrt{8-3x^2} + x}{\sqrt{8-3y^2 + y}}. $$
This result doesn't seem to help me as it is ultimately telling me something very trivial that I already know about the relationship between x and y coordinates and the tan function.
I know that my choice of coordinates might not be particularly strategic. Are there any other mistakes I'm making in how I'm approaching this?
Thank you!