The points of contact of the tangents drawn from the origin to the curve $\mathrm{y}={\mathrm{x}}^2+3\mathrm{x}+4$ are

Point on the Curve $\mathrm{P}\left({\mathrm{x}}_1 {\mathrm{y}}_1\right)$ is ${\mathrm{y}}_1={{\mathrm{x}}_1}^2+3{\mathrm{x}}_1+4 -\left(1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x}+3 \Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{P}\left({\mathrm{x}}_1 {\mathrm{y}}_1\right)} =2{\mathrm{x}}_1+3$

eq of Tangent $\mathrm{y}-{\mathrm{y}}_1=\left(2{\mathrm{x}}_1+3\right) \left(\mathrm{x}-{\mathrm{x}}_1\right)$

Passes through $\left(0, 0\right) \Rightarrow -{\mathrm{y}}_1=\left(2{\mathrm{x}}_1+3\right) \left(-{\mathrm{x}}_1\right)$

eq (1) $\mathrm{sub} \Rightarrow {{\mathrm{x}}_1}^2+3{\mathrm{x}}_1+4={\mathrm{x}}_1\left(2{\mathrm{x}}_1+3\right)$

                    $\begin{array}{rcl} & \Rightarrow & {{\mathrm{x}}_1}^2=4\\ & \Rightarrow & {\mathrm{x}}_1=\pm 2\\ \mathrm{x}1& =& 2 \Rightarrow {\mathrm{y}}_1=4+6+4=14\\ {\mathrm{x}}_1& =& -2 \Rightarrow {\mathrm{y}}_1=4-6+4=2\end{array}$

$\therefore$ Points $(2, 14) (-2, 3)$