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Can every manifold that is locally given as the solution to quadratic equations also be locally represented as the graph of a quadratic function?

If not, is this true for a nice subclass of manifolds? Is there an easy counterexample or a reference for this?

My setting of interest in this is mainly $\mathcal{C}^2$ submanifolds of $\mathbb{R}^n$.

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An ellipse is locally represented by solutions to an equations that's quadratic in each of $x$ and $y$, but is nowhere represented as the graph of a quadratic. But maybe that's not what you meant.

A tilted parabola (e.g., $(y+x)^2 = y - x$) is locally represented by a solution to a quadratic (in $x + y$) but is nowhere the graph of a quadratic in either of $x$ or $y$.

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    Thanks, this actually is what I want. However, is there a way to really see that this cannot be a quadratic graph locally in either $x$ or $y$? – 2017-01-15
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    @MaxM: yes, just solve for $x$ or $y$ and observe it is not a quadratic function. For example, it is not differentiable on $\mathbb R$. – 2017-01-16
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    Sure. Look at $x = 0$: at $x = 0$, both $y = 1$ and $y = -1$ are on the graph. Similarly, for $y = 0$, both $x = 0$ and $x = -1$ are on the graph. – 2017-01-16