The locus of the point of intersection of the straight lines $\mathrm{tx}–2\mathrm{y}–3\mathrm{t}=0; \mathrm{x}–2\mathrm{ty}+3=0 \left(\mathrm{t}\in \mathrm{R}\right)$ is

$$\begin{gathered} \mathrm{Given} \mathrm{lines} \mathrm{t}(\mathrm{x}-3)=2\mathrm{y} \\ \Rightarrow \mathrm{t}=\frac{2\mathrm{y}}{\mathrm{x}-3} — ① \\ and \mathrm{x}-2\mathrm{ty}+3=0 \Rightarrow \mathrm{t}=\frac{\mathrm{x}+3}{2\mathrm{y}} — ② \\ \mathrm{them} ① \& ① \frac{2\mathrm{y}}{\mathrm{x}-3}=\frac{\mathrm{x}+3}{2\mathrm{y}} \\ \Rightarrow 4{\mathrm{y}}^2={\mathrm{x}}^2-9 \\ \Rightarrow {\mathrm{x}}^2-4{\mathrm{y}}^2=9 \\ \Rightarrow \frac{x^2}{9}-\frac{y^2}{{\displaystyle \frac{9}{4}}}=1 \\ It represents hyperbola \\ \therefore \mathrm{Length} \mathrm{of} \mathrm{conju}gate \mathrm{axis} =2\mathrm{b} \\ =2\left(\frac{3}{2}\right) \\ =3 \end{gathered}$$