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**Hint:**As we are given two straight lines we find the value of t from (2) and using this in equation (1) we get a equation of a conic and comparing it with the general equation of ellipse and hyperbola we get that it is a hyperbola with ${a^2} = 9$ and ${b^2} = \dfrac{9}{4}$ and now we can find the eccentricity using the formula $\sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} $ and conjugate axis is 2b and see which matches the given options.

**Complete step by step solution:**

We are given two straight lines

$ \Rightarrow tx - 2y - 3t = 0$ ………(1)

$ \Rightarrow x - 2ty + 3 = 0$……….(2)

From the second equation we can find the value of t

$

\Rightarrow x + 3 = 2ty \\

\Rightarrow \dfrac{{x + 3}}{{2y}} = t \\

$

Using the value of t in equation (1)

$

\Rightarrow \left( {\dfrac{{x + 3}}{{2y}}} \right)x - 2y - 3\left( {\dfrac{{x + 3}}{{2y}}} \right) = 0 \\

\Rightarrow \left( {\dfrac{{{x^2} + 3x}}{{2y}}} \right) - 2y - \left( {\dfrac{{3x + 9}}{{2y}}} \right) = 0 \\

\Rightarrow \dfrac{{{x^2} + 3x - 4{y^2} - 3x - 9}}{{2y}} = 0 \\

\Rightarrow \dfrac{{{x^2} - 4{y^2} - 9}}{{2y}} = 0 \\

$

Cross multiplying and taking the constant to the other side we get

$

\Rightarrow {x^2} - 4{y^2} - 9 = 0 \\

\Rightarrow {x^2} - 4{y^2} = 9 \\

$

In the given options we are given that its either a ellipse or hyperbola

Hence to bring it to that form lets divide throughout by 9

$

\Rightarrow \dfrac{{{x^2}}}{9} - \dfrac{{4{y^2}}}{9} = \dfrac{9}{9} \\

\Rightarrow \dfrac{{{x^2}}}{9} - \dfrac{{4{y^2}}}{9} = 1 \\

$

We know that the general form of the ellipse and hyperbola are $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ and }}\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ }}$ respectively

Hence from this we get that the locus is a hyperbola

In our option we have a hyperbola with eccentricity $\sqrt 5 $ or with a length of conjugate axis 3

So now lets find the eccentricity of our hyperbola and its conjugate axis

In a hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ }}$the eccentricity is given by $\sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} $ and the conjugate axis is given as 2b

Now let's find the eccentricity

Here ${a^2} = 9$ and ${b^2} = \dfrac{9}{4}$

Using this we get

$

\Rightarrow e = \sqrt {1 + \dfrac{{\dfrac{9}{4}}}{9}} \\

\Rightarrow e = \sqrt {1 + \dfrac{1}{4}} \\

\Rightarrow e = \sqrt {\dfrac{{4 + 1}}{4}} = \sqrt {\dfrac{5}{4}} = \dfrac{{\sqrt 5 }}{2} \\

$

Here we get the eccentricity to be $\dfrac{{\sqrt 5 }}{2}$which does not match the given option

So now lets find the conjugate axis

The conjugate axis is 2b

Since ${b^2} = \dfrac{9}{4}$we get$b = \sqrt {\dfrac{9}{4}} = \dfrac{3}{2}$

Hence our conjugate axis is

$ \Rightarrow 2b = 2\left( {\dfrac{3}{2}} \right) = 3$

From this we get that the locus is a hyperbola with conjugate axis 3

**Therefore the correct answer is option D.**

**Note :**

A hyperbola is created when the plane intersects both halves of a double cone, creating two curves that look exactly like each other, but open in opposite directions.

The eccentricity of a hyperbola is greater than 1. This indicates that the distance between a point on a conic section the nearest directrix is less than the distance between that point and the focus.

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