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Thread: Cubics

  1. #1
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    Cubics

    (2.1) Examples of parametrised cubics. Some plane cubic curves can be parametrised, just as the conics:

    Nodal cubic $\displaystyle C: (y^2=x^3+x^2) \subset \mathbb{R}^2$ is the image of the map $\displaystyle \varphi: \mathbb{R}^1 \to \mathbb{R}^2$ given by $\displaystyle t\mapsto (t^2-1, t^3-t)$

    Cuspidal cubic. $\displaystyle C: (y^2 = x^3)\subset \mathbb{R}^2$ is the image of $\displaystyle \varphi:\mathbb{R}^1 \to \mathbb{R}^2$ given by $\displaystyle t\mapsto (t^2,t^3)$


    (1) Let $\displaystyle C: (y^2 = x^3 + x^2) \subset \mathbb{R}^2.$ Show that a varible line though $\displaystyle (0,0)$ meets $\displaystyle C$ at one further point, and hence deduce the parametrisation of $\displaystyle C$ given in (2.1). Do the same for $\displaystyle (y^2 = x^3)$ and $\displaystyle (x^3 = y^3-y^4).$

    If anyone could show me how to do any of those in (1), I would really appreciate it.. I have no idea how to tackle this problem. Thanks!
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  2. #2
    MHF Contributor
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    The first step is to determine what a line through the origin looks like.
    The equation of this line will be y = kx for some k. Then:

    $\displaystyle y^2 = x^3 + x^2$

    Substituting for y:

    $\displaystyle (kx)^2 = x^3 + x^2$

    $\displaystyle x^3 = (kx)^2 - x^2$

    $\displaystyle x^3 = k^2x^2 - x^2$

    $\displaystyle x^3 = (k^2 - 1)x^2$

    For $\displaystyle x \neq 0$:

    $\displaystyle x = k^2 - 1$

    And since $\displaystyle y = kx$:

    $\displaystyle y = k(k^2 - 1)$
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