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Thread: Proof of Conics in R^2

  1. #1
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    Proof of Conics in R^2

    I really suck at proofs and I have this due tomorrow...any help would be appreciated. It doesn't count toward any points but I need to turn it in for credit

    Definition: (1.3) Conics in $\displaystyle \mathbb{R}^2$. A conic in $\displaystyle \mathbb{R}^2$ is a plane curve given by the quadratic equation $\displaystyle q(x,y) = ax^2 + bxy +cy^2 + dx +ey +f = 0$

    Here is my question:
    Prove the statement in (1.3) that an affine transformation can be used to put any conic of $\displaystyle \mathbb{R}^2$ into one of the standard forms (a-1). (Hint: use a linear transformation $\displaystyle x \to Ax$ to take the leading term $\displaystyle ax^2 + by + cy^2$ into one of $\displaystyle \underline{+}x^2\underline{+}y^2$ or $\displaystyle \underline{+}x^2$ or $\displaystyle 0$; then complete the square in $\displaystyle x$ and $\displaystyle y$ to get rid of as much of the linear part as possible.

    Thank you for your time and consideration. I will be up for awhile tonight to answer any questions and try to figure this out!
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by shadow_2145 View Post
    I really suck at proofs and I have this due tomorrow...any help would be appreciated. It doesn't count toward any points but I need to turn it in for credit

    Definition: (1.3) Conics in $\displaystyle \mathbb{R}^2$. A conic in $\displaystyle \mathbb{R}^2$ is a plane curve given by the quadratic equation $\displaystyle q(x,y) = ax^2 + bxy +cy^2 + dx +ey +f = 0$

    Here is my question:
    Prove the statement in (1.3) that an affine transformation can be used to put any conic of $\displaystyle \mathbb{R}^2$ into one of the standard forms (a-1). (Hint: use a linear transformation $\displaystyle x \to Ax$ to take the leading term $\displaystyle ax^2 + by + cy^2$ into one of $\displaystyle \underline{+}x^2\underline{+}y^2$ or $\displaystyle \underline{+}x^2$ or $\displaystyle 0$; then complete the square in $\displaystyle x$ and $\displaystyle y$ to get rid of as much of the linear part as possible.

    Thank you for your time and consideration. I will be up for awhile tonight to answer any questions and try to figure this out!
    well, i'll do this in the projective plane, that is you can always have a transformation from the affine plane to the projective plane.. (i'll just copy the notes i have here)

    let $\displaystyle F(x,y,z) = ax^2 + bxy +cy^2 + dxz +eyz +fz^2$

    case1: $\displaystyle a=c=f=0$
    at least one of $\displaystyle b,d,e$ is not 0.
    WLOG, let $\displaystyle b\not= 0$.
    consider the transformation:
    $\displaystyle x' = x$
    $\displaystyle y' = -x+y$
    $\displaystyle z'=z$

    arrive at:...

    case2: at least one of $\displaystyle a,c,f$ is not 0.
    WLOG, let $\displaystyle a\not=0$
    and another WLOG, let $\displaystyle a=1$.
    $\displaystyle F(x,y,z) = x^2 + bxy +cy^2 + dxz +eyz +fz^2$
    $\displaystyle = \left(x +\frac{b}{2}y+\frac{d}{2}z\right)^2 + \left(c-\frac{b^2}{4}\right)y^2 + \left(e-\frac{bd}{2}\right)yz + \left(f-\frac{d^2}{4}\right)z^2$

    consider:
    $\displaystyle x'=x + \frac{b}{2}y + \frac{d}{2}z \Rightarrow x = x' -\frac{b}{2}y'-\frac{d}{2}z'$
    $\displaystyle y'=y$
    $\displaystyle z'=z$

    arrive at: ...
    WLOG, assume $\displaystyle b=0,d=0$
    left with: $\displaystyle F(x,y,z) = x^2 +cy^2+eyz+fz^2$

    case3: from $\displaystyle F(x,y,z) = x^2 +cy^2+eyz+fz^2$
    if $\displaystyle c=e=f=0$
    $\displaystyle F(x,y,z)=x^2$

    case4: from $\displaystyle F(x,y,z) = x^2 +cy^2+eyz+fz^2$, at least one of $\displaystyle c,e,f$ is not = 0
    if $\displaystyle e\not=0$ and $\displaystyle c=0,f=0$
    $\displaystyle F(x,y,z)=x^2 + eyz$

    consider:
    $\displaystyle x'=x$
    $\displaystyle y'=y-z$
    $\displaystyle z'=z$

    arrive at:
    $\displaystyle F(x',y',z') = x'^2 + e(y'+z')z' = x'^2 + ey'z' + ez'^2$ or
    $\displaystyle F(x,y,z) = x^2 + eyz + ez^2$


    If $\displaystyle c\not=0$, by interchanging $\displaystyle y$ and $\displaystyle z$, we can force $\displaystyle f\not=0$.

    now, WLOG, let $\displaystyle f\not=0$

    case5: $\displaystyle F(x,y,z) = x^2 +cy^2+eyz+fz^2$, $\displaystyle f\not=0$

    let $\displaystyle t = \sqrt{|f|}$, $\displaystyle t^2 = \pm f$

    consider $\displaystyle \varphi$:
    $\displaystyle x'=x$
    $\displaystyle y'=y$
    $\displaystyle z'=tz$

    by this transformation, you shall arrive at:
    $\displaystyle F^{\varphi}(x,y,z) = x^2 + \left(c \mp \frac{e^2}{4t^2}\right)y^2 \pm \left(\frac{e}{2t}y+z\right)^2$

    consider $\displaystyle \psi$:
    $\displaystyle x'=x$
    $\displaystyle y'=y$
    $\displaystyle z'=\frac{e}{2t}y+z$

    arrive at:
    $\displaystyle (F^{\varphi})^{\psi}=F^{\psi\varphi} = x^2 + \left(c \mp\frac{e^2}{4t^2}\right)y^2 \pm z^2$

    case6:
    $\displaystyle F(x,y,z) = x^2 + Cy^2 \pm z^2$, $\displaystyle C\not=0$

    set $\displaystyle s = \sqrt{|C|}$
    consider:
    $\displaystyle x'=x$
    $\displaystyle y'=sy$
    $\displaystyle z'=z$

    you arrive at: $\displaystyle F(x,y,z) =x^2 \pm y^2 \pm z^2$

    so from there, consider all the combinations of plus and minus there..

    case7: if $\displaystyle C=0$

    $\displaystyle F(x,y,z)= x^2 \pm z^2$

    consider:
    $\displaystyle x'=x$
    $\displaystyle y'=z$
    $\displaystyle z'=y$

    you arrive at: $\displaystyle F(x',y',z')=x'^2 \pm y'^2$ or simply $\displaystyle F(x,y,z)=x^2 \pm y^2$


    -------------------end-------------------

    well, the discussion is too long so try to understand it by yourself.. and this is just a guide or outline of the whole proof..

    and another: the reason why there are 7 cases is that, this actually shows that any second degree polynomial in the affine plane can be transformed into one of the following:
    1. empty set
    2. single point
    3. a doubled line i.e. two lines which are coinciding
    4. two distinct lines
    5. a conic..

    you shall see which cases gives you the conics (which you are looking for)..
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