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How to find equations of two lines given a conic equation?

  1. Apr 13, 2012 #1
    I came accross the question
    The following equation represents two straight lines. Determine the equation of each of the two lines.
    x^2+8 xy−x+16 y^2−4 y−72=0

    I know we need to take common factors and the answer is
    (x+4 y-9) (x+4 y+8) = 0
    leading to equations y = 1/4 (-x-8) and y = (9-x)/4 but how did we get to (x+4 y-9) (x+4 y+8) = 0 in the first place?? Thanks
  2. jcsd
  3. Apr 13, 2012 #2


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    Science Advisor

    My first thought with any such conic, not knowing in advance whether it represents lines or not, would be to put it in "standard form" by eliminating the "xy" term.

    Since an "xy" term represents a rotation of the axes, we can do that in either of two (equivalent) ways.

    1) Write [itex]x= x'cos(\theta)- y'sin(\theta)[/itex], [itex]y= x' sin(\theta)+ y' cos(\theta)[/itex]. Then [itex]x^2= x'^2cos^2(\theta)- 2x'y' sin(\theta)cos(\theta)+ y'^2sin^2(\theta)[/itex], [itex]xy= x'^2 sin(\theta)cos(\theta)+ x'y'(cos^2(\theta)- sin^2(\theta))+ y'^2sin(\theta)cos(\theta)[/itex], and [itex]y^2= x'^2 sin^2(\theta)+ 2x'y' sin(\theta)cos(\theta)+ y'^2 cos^2(\theta)[/itex]. Put those into the equation and choose [itex]\theta[/itex] so that the coefficient of the x'y' term is 0.

    2) More sophisticated but but simpler calculations: write the quadratic term, [itex]x^2+ 8xy+ 16y^2[/itex] as a matrix multiplication:
    [tex]\begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}1 & 4 \\ 4 & 16\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}[/tex]
    where I have "distributed" the coefficient of xy between the two "anti-diagonal" terms of the matrix in order to get a symmetric matrix so that it will have two independent eigenvalues.

    Now, find the eigenvalues and eigenvectors of that matrix:
    [tex]\left|\begin{array}{cc}1- \lambda & 4 \\ 4 & 16-\lambda\end{array}\right|= \lambda^2- 17\lambda= 0[/tex]
    which has solutions [itex]\lambda= 0[/itex] and [itex]\lambda= 17[/itex]. The fact that one eigenvalue is 0 signals "parabola" (both the same sign, ellipse, differing signs, hyperbola) but "two parallel straight lines" is a special case of "parabola".

    We look for eigenvectors corresponding to those eigenvalues. If <x y> is an eigenvector corresponding to eigenvalue 0, we must have
    [tex]\begin{bmatrix}1 & 4 \\ 4 & 16\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}x+ 4y \\ 4x+ 16y\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/tex]
    which gives the two equations x+ 4y= 0 and 4x+ 16y= 0 both of which reduce to x= -4y or x+ 4y= 0. That is the line in the direction of the eigenvector <-4, 1>.

    For the eigenvector corresponding to eigenvalue 17 we have
    [tex]\begin{bmatrix}1 & 4 \\ 4 & 16\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}x+ 4y \\ 4x+ 16y\end{bmatrix}= \begin{bmatrix}17x \\ 17y\end{bmatrix}[/tex]
    which gives the equations x+ 4y= 17x and 4x+ 16y= 17y both of which reduce to y= 4x which is the same as y- 4x= 0. That is a line in the direction of the eigenvector <1 4>.

    What that says is we can get rid of the xy term by using those lines as axes- that is define x'= x+ 4y and y'= y- 4x which are the same as x= (1/17)x'+ (4/17)y' and y= (4/17)x'+ (1/17)y'. Replace x and y in the original conic to eliminate the xy term.
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