Consider the rational normal curve of degree 4. It can be realized as the intersection of a certain number of quadrics. Usually in examples we consider those quadrics to be singular. However, can at least two of those quadrics be non-singular?
Edit: I mainly wish to know if it is possible for the rational normal curve of degree 4 to lie on the intersection of two non-singular quadrics.
The rational normal curve
\begin{array}{rcl} \mathbb{P}^1 & \to & \mathbb{P}^4 \\ [s:t] & \mapsto & [s^4 : s^3 t : s^2 t^2 : s t^3 : t^4 ] \end{array}
is cut out of $\mathbb{P}^4$ by the quadrics
$$x_0 x_2 - x_1^2 = x_1 x_3 - x_2^2 = x_2 x_4 - x_3^2 = x_0 x_4 - x_1 x_3 = 0$$
A quadric hypersurface is singular iff the corresponding quadric form is not of full rank.
So write
$$A = \left( \begin{array}{ccccc} 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right), \qquad B = \left( \begin{array}{ccccc} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & 0 \\ 0 & 0 & -1 & 0 & 0 \\ 0 & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right),$$ $$C = \left( \begin{array}{ccccc} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & \frac{1}{2} & 0 & 0 \\ \end{array} \right), \qquad D = \left( \begin{array}{ccccc} 0 & 0 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & 0 & 0 \\ \end{array} \right)$$
and pick some linear combinations making the determinants non-vanishing, e.g. $A + B - D$.
Proceeding in this way, you can see e.g. that the rational normal curve of degree 4 is the intersection of the four smooth quadrics
$$-x_1^2 + x_0 x_2 - x_2^2 - x_1 x_3 + 2 x_0 x_4 = 0,$$ $$-x_1^2 + x_0 x_2 - x_2^2 + 2 x_1 x_3 - x_0 x_4 = 0,$$ $$-x_1^2 + x_0 x_2 - x_1 x_3 - x_3^2 + x_0 x_4 + x_2 x_4 = 0,$$ $$-x_2^2 + 2 x_1 x_3 - x_3^2 - x_0 x_4 + x_2 x_4 = 0.$$
Of course, we don't expect all these intersections to be nice, especially since this isn't a complete intersection -- we're cutting a one-dimensional object out of a four-dimensional space, and using four (i.e., "one too many") equations to do it.