The quadric $Q$ is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$ embedded with the linear system $\mathcal{O}(1,1)$. \mathcal{O}_Q(1,1)$. The rulings of the quadric are the two families of lines $\mathbb{P}^1 \times \{a\}$ and $\{b\} \times \mathbb{P}^1$, and it is not difficult to see that these are the only lines in$Q$. It is also known that the locus of lines meeting a given line$\ell \subset \mathbb{P}^3$is a hyperplane section under via the Plucker Plücker embedding$G(2,4) \subset \mathbb{P}^5$(see Harris, Algebraic Geometry, p. 244). Now by Bézout theorem a fixed line$\ell$intersects$Q$in two points, and for each point there is precisely a line of the first ruling and a line of the second ruling. So a hyperplane section of the Grasmannian intersects the locus corresponding to the first ruling in two points, and similarly for the second ruling. It follows that the locus corresponding to the first ruling via the Plücker embedding has degree$2$under the Plucker embedding. in$\mathbb{P}^5$. On the other hand, such a locus must be a curve isomorphic to$\mathbb{P}^1$, since a ruling is a$1$-dimensional linear system in$Q$; so it is a smooth conic. The same is obviously clearly true for the locus corresponding to the second ruling. Finally, no line belongs to both rulings. Then the locus of lines of$Q$corresponds, via the Plucker Plücker embedding$G(2,4) \subset \mathbb{P}^5$, to the disjoint union of two smooth conics. 3 added 328 characters in body The quadric$Q$is isomorphic to$\mathbb{P}^1 \times \mathbb{P}^1$embedded with the linear system$\mathcal{O}(1,1)$. The rulings of the quadric are the two families of lines $\mathbb{P}^1 \times \{a\}$ and $\{b\} \times \mathbb{P}^1$, and it is not difficult to see that these are the only lines in$Q$. It is also known that the locus of lines meeting a given line$\ell \subset \mathbb{P}^3$is a hyperplane section under the Plucker embedding$G(2,4) \subset \mathbb{P}^5$(see Harris, Algebraic Geometry, p. 244). Now by Bézout theorem a fixed line$\ell$intersects the quadric$Q$in two points, and for each point there is precisely a line of the first ruling and a line of the second ruling. So a hyperplane section of the Grasmannian intersects the locus corresponding to the first ruling in two points, and similarly for the second ruling. It follows that the locus corresponding to the first ruling has degree$2$under the Plucker embedding. On the other hand, such a locus must be a curve isomorphic to$\mathbb{P}^1$, since a ruling is a$1$-dimensional linear system in$Q$; so it is a smooth conic. The same is obviously true for the locus corresponding to the second ruling. Finally, no line belongs to both rulings. Then the locus of lines of$Q$corresponds, via the Plucker embedding$G(2,4) \subset \mathbb{P}^5$, to the disjoint union of two smooth conics. 2 added 1 characters in body It is known that the locus of lines meeting a given line$\ell \subset \mathbb{P}^3$is a hyperplane section under the Plucker embedding$G(2,4) \subset \mathbb{P}^5$(see Harris, Algebraic Geometry, p. 244). Now a fixed line$\ell$intersects the quadric$Q$in two points, and for each point there is precisely a line of the first ruling and a line of the second ruling. So a hyperplane section of the Grasmannian intersect intersects the locus corresponding to the first ruling in two points, and similarly for the second ruling. It follows that the locus corresponding to the first ruling has degree$2$under the Plucker embedding. On the other hand, such a locus must be a curve isomorphic to$\mathbb{P}^1$, since a ruling is a$1$-dimensional linear system in$Q$; so it is a smooth conic. The same is obviously true for the locus corresponding to the second ruling. Finally, no line belongs to both rulings. Then the locus of lines of$Q$corresponds, via the Plucker embedding$G(2,4) \subset \mathbb{P}^5\$, to the disjoint union of two smooth conics.