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 $y$ ?

  • $\begingroup$ Maybe this would help math.stackexchange.com/q/1011192/463578 $\endgroup$ Aug 28, 2022 at 8:20
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    $\begingroup$ Please correct "eclipse" to "ellipse". And try to use MathJax in formulas. $\endgroup$ Aug 28, 2022 at 10:17
  • $\begingroup$ 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. $\endgroup$ Aug 28, 2022 at 10:25

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – GEdgar
    Aug 28, 2022 at 11:55

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