# solution of x and y for non standard eclipse equation

with my basic math knowledge and search I found out that for a standard eclipse equation $$x^2 / a^2 + y^2/b^2 = 1$$, we can find out $$x$$ and $$y$$ by setting $$y$$ and $$x$$ intercept $$0$$. But if the equation is of the form $$x^2 + xy + 41y^2 = n$$, how do we make a standard formula for $$x$$ and $$y$$ ?

• Maybe this would help math.stackexchange.com/q/1011192/463578 Aug 28, 2022 at 8:20
• Please correct "eclipse" to "ellipse". And try to use MathJax in formulas. Aug 28, 2022 at 10:17
• To compute the semi-axes of the ellipse, consider a generic line through the origin $y=mx$ and find its intersections $(x(m), y(m))$ with the ellipse as a function of $m$: minimum and maximum of $x(m)^2+y(m)^2$ will give you the squares of the semi-axes. Aug 28, 2022 at 10:25

Note that in a more general form, your equation is of the form: $${\alpha _1}{x_1}^2 + 2{\alpha _2}{x_1}{x_2} + {\alpha _3}{x_2}^2 = {\alpha _0}$$ which using the matrix notion, may be written as: $$\left[ {\begin{array}{*{20}{c}} {{x_1}}&{{x_2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{\alpha _1}}&{{\alpha _2}} \\ {{\alpha _2}}&{{\alpha _3}} \end{array}} \right]\left[ \begin{gathered} {x_1} \hfill \\ {x_2} \hfill \\ \end{gathered} \right] = {\alpha _0}$$ Now, assuming a general linear transformation of the form $$Y = LX$$, where $$X = \left[ \begin{gathered} {x_1} \hfill \\ {x_2} \hfill \\ \end{gathered} \right]$$, one may re-write the original equation as: $${Y^{\text{T}}}{\left( {{L^{ - 1}}} \right)^{\text{T}}}\left[ {\begin{array}{*{20}{c}} {{\alpha _1}}&{{\alpha _2}} \\ {{\alpha _2}}&{{\alpha _3}} \end{array}} \right]{L^{ - 1}}Y = {\alpha _0}$$ Using eigen-decomposition, the symmetric coefficient matrix could be re-written as: $$\left[ {\begin{array}{*{20}{c}} {{\alpha _1}}&{{\alpha _2}} \\ {{\alpha _2}}&{{\alpha _3}} \end{array}} \right] = PD{P^{\text{T}}}$$ where $$D$$ is a diagonal matrix. Substituting in the equation on has: $${Y^{\text{T}}}{\left( {{L^{ - 1}}} \right)^{\text{T}}}PD{P^{\text{T}}}{L^{ - 1}}Y = {\alpha _0}$$ By factoring, one may write the above as: $$\left( {{P^{\text{T}}}{L^{ - 1}}Y} \right)D\left( {{P^{\text{T}}}{L^{ - 1}}Y} \right) = {\alpha _0}$$ Since in the final form, no cross terms (e.g. $${y_1}{y_2}$$) are allowed, one must enforce: $${P^{\text{T}}}{L^{ - 1}} = I$$ which leads to: $$L = {P^{\text{T}}}$$ So in summary, you should decompose your coefficients matrix, and the linear transformation matrix which is simply $${P^{\text{T}}}$$ would transform your equation into your desired canonical form.
• An older way of saying this: rotate the coordinates. In $x^2 + xy + 41y^2 = n$, the axes of the ellipse are slightly tilted with respect to the $x,y$ axes. Aug 28, 2022 at 11:55