Hints

Let C be a smooth curve of degree d on a quadric cone Q. Then every ruling of Q intersects C in the same number of points (= d/2 if d is even, (d+1)/2 if d is odd).

To see it, consider 2 rulings, r, r', and assume r (resp. r') intersects C in a (resp. a') points, where a > a'. Take a third ruling, r''. The ruling r" intersects C in b points. The plane spanned by r and r'' intersects C in a+b = d points or in d = a+b-1 points (if the vertex lies on both r and r'').The plane spanned by r' and r'' intersects C in a'+b = d points or in d = a'+b-1 points (if the vertex lies on both r' and r''). The only possibility would be: a-1 = a', the vertex lies on r and r'' but not on r', which is absurd.

Through any point of C passes a trisecant to C.

(To prove it, consider the projection on a plane from a point, p, of C: if C has no 3secants through p, you get a smooth plane curve of degree 3, genus 0: impossible.)

Use Riemann-Roch to compute h^0(O_C(2)) = 8; it follows that h^0(I_C(2)) >= 2. Conclude with Bezout.

The equation of a quadric in P3 is an homogeneous polynomial of degree 2 in 4 variables, i.e. a quadratic form; consider the symmetric 4x4 matrix associated to this quadratic form; the quadric is singular iff the determinant of this matrix is zero.

Consider the determinant of the matrix associated to uQ+vQ' as a (homogeneous) polynomial in u, v.

Let H, H' be two planes in a pencil (H, H' through a same line) and take a third plane, H", not in the pencil. Set Q = H.H", Q' = H'.H". Every quadric in the pencil generated by Q and Q' is of the form H".(uH+vH'), hence it is singular. In this case, the determinant of the matrix associated to uQ+vQ', viewed as a polynomial in u, v, is identically zero.