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I am studying Differential geometry I tried to prove this by taking all three components as quadratic with $t$ as a parameter but could not be successful.

If all three component functions of a space curve $\gamma$ are quadratic functions, prove that the image of $\gamma$ is contained in a plane.

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HINT: The vector space of polynomials of degree $\le 2$ in $t$ is $3$-dimensional, so the vector space of polynomials of degree $\le 2$ that vanish at $t=0$ is $2$-dimensional. If $\gamma(t)$ is your curve, consider the curve $\gamma(t)-\gamma(0)$.

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