How do you write an equation for the hyperbola with center at $\left( { - 2, - 3} \right)$, focus at $\left( { - 4, - 3} \right)$, vertex at $\left( { - 3, - 3} \right)$?

Hint: In this particular question use the concept that the center of the hyperbola is at the midpoint of the line joining the two vertices of the hyperbola and the distance between the vertices is the length of the major axis i.e. 2a, and use that the coordinates of the focus from the center are at (ae, 0), where e is the eccentricity, so use these concepts to solve the equation.

Complete step-by-step solution:
Given data:
The vertex of a hyperbola is at \[\left( { - 3, - 3} \right)\].
One focus of the hyperbola is at $\left( { - 4, - 3} \right)$.
The Center of the hyperbola is $\left( { - 2, - 3} \right)$.
As we all know that the general equation of the hyperbola is given as,
$\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1$
where (h, k) is the center of the hyperbola and a, and b are the length of the semi-major and semi-minor axis of the hyperbola.
Therefore, \[h = - 2\] and $k = - 3$.
As the vertex is at $\left( { - 3, - 3} \right)$, so the distance between the vertex and center is 1.
$ \Rightarrow a = 1$
Now it is given that one of its focus is at $\left( { - 4, - 3} \right)$.
So, the distance of the focus from the center $\left( { - 2, - 3} \right)$ is
$ \Rightarrow $ Distance of focus $ = - 2 - \left( { - 4} \right)$
Open the bracket,
$ \Rightarrow $ Distance of focus $ = - 2 + 4$
Simplify the terms,
$ \Rightarrow $ Distance of focus $ = 2$
Now the coordinates of the focus from the center are at \[\left( {ae,0} \right)\], where e is the eccentricity.
Therefore, $ae = 2$
Substitute the values of a,
$ \Rightarrow 1 \times e = 2$
Simplify the terms,
$ \Rightarrow e = 2$
Now as we all know that in hyperbola the relation between a, b and e is
${b^2} = {a^2}\left( {{e^2} - 1} \right)$
Substitute the values,
$ \Rightarrow {b^2} = {1^2}\left( {{2^2} - 1} \right)$
Simplify the terms,
$ \Rightarrow {b^2} = 4 - 1$
Subtract the values,
$ \Rightarrow {b^2} = 3$
So, the equation of the hyperbola becomes,
$ \Rightarrow \dfrac{{{{\left( {x - \left( { - 2} \right)} \right)}^2}}}{{{1^2}}} - \dfrac{{{{\left( {y - \left( { - 3} \right)} \right)}^2}}}{3} = 1$
Simplify the terms,
$ \Rightarrow \dfrac{{{{\left( {x + 2} \right)}^2}}}{1} - \dfrac{{{{\left( {y + 3} \right)}^2}}}{3} = 1$

Hence, the required hyperbola is $\dfrac{{{{\left( {x + 2} \right)}^2}}}{1} - \dfrac{{{{\left( {y + 3} \right)}^2}}}{3} = 1$.

Note: Whenever we face such types of questions the key concept, we have to remember is always recall that the general equation of the hyperbola is given as $\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1$, where (h, k) is the center of the hyperbola and a, and b are the length of the semi-major and semi-minor axis of the hyperbola and in hyperbola the relation between a, b and e is ${b^2} = {a^2}\left( {{e^2} - 1} \right)$.