1. Given family of lines

We need the locus of the intersection point of $tx-2y-3t=0 \quad ...(1)$ $x-2ty+3=0 \quad ...(2)$ where $t\in\mathbb R$.

Let the intersection point be $(x,y)$.

2. Express $t$ from both equations

From (1): $tx-2y-3t=0$ $t(x-3)=2y$ $t=\frac{2y}{x-3} \quad ...(3)$ provided $x\ne 3$.

From (2): $x-2ty+3=0$ $2ty=x+3$ $t=\frac{x+3}{2y} \quad ...(4)$ provided $y\ne 0$.

Equating (3) and (4): $\frac{2y}{x-3}=\frac{x+3}{2y}$

Cross-multiplying, $4y^2=(x-3)(x+3)=x^2-9$

So, $x^2-4y^2=9$

or $\frac{x^2}{9}-\frac{y^2}{9/4}=1$

3. Identify the conic

This is a hyperbola of standard form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with $a^2=9,\quad b^2=\frac94$

Hence, $a=3,\quad b=\frac32$

4. Check the options

Option A: ellipse with eccentricity $\dfrac{2}{\sqrt5}$

False, because the locus is a hyperbola, not an ellipse.

Option B: ellipse with the length of major axis 6

False, again not an ellipse.

Option C: hyperbola with eccentricity $\sqrt5$

For the hyperbola, $c^2=a^2+b^2=9+\frac94=\frac{45}{4}$ $c=\frac{3\sqrt5}{2}$ So eccentricity is $e=\frac{c}{a}=\frac{(3\sqrt5/2)}{3}=\frac{\sqrt5}{2}$ Not $\sqrt5$. So this is false.

Option D: hyperbola with the length of conjugate axis 3

For hyperbola, length of conjugate axis is $2b=2\cdot \frac32=3$ So this is true.

5. Final answer

The locus is $x^2-4y^2=9$ which is a hyperbola whose conjugate axis has length $3$.

Therefore, the correct option is D.