Find an equation of a hyperbola with vertices (0​,3) and ​(−4​,3) and e=3/2.

Thank you

2a = 4

a = 2

e = c/a
c/2 = 3/2
c = 3

in a hyperbola:
a^2 + b^2 = c^2
4 + b^2 = 9
b^2 = 5
a^2 = 4
centre is (-2,3)

(x+2)^2 /4 - (y-3)^2 /5 = 1

To find the equation of a hyperbola, we can use the standard form equation for hyperbolas with horizontal transverse axis:

((x-h)^2)/a^2 - ((y-k)^2)/b^2 = 1

First, we need to find the center of the hyperbola. It is the midpoint between the vertices, which can be calculated using the midpoint formula:

Midpoint formula: ( (x1+x2)/2 , (y1+y2)/2 )

The vertices are (0, 3) and (-4, 3). Plugging these values into the midpoint formula, we get:
( (0 + (-4))/2 , (3 + 3)/2 ) = ( -2 , 3 ).

So, the center of the hyperbola is (-2, 3).

Next, we need to find the value of a, which is the distance from the center to either vertex. Using the distance formula:

Distance formula: sqrt( (x2-x1)^2 + (y2-y1)^2 )

Plugging in the values, we get:
sqrt( (-4 - (-2))^2 + (3-3)^2 ) = sqrt( (-4 + 2)^2 + 0^2 ) = sqrt( (-2)^2 + 0 ) = sqrt( 4 + 0 ) = sqrt( 4 ) = 2.

So, the value of a is 2.

We are also given e = 3/2. We can use the relationship between a, b, and e to find the value of b.

e = c/a, where c is the distance from the center to either focus.

In a hyperbola, c is related to a and b by the equation:
c^2 = a^2 + b^2

Substituting the values given:
(3/2)^2 = 2^2 + b^2
9/4 = 4 + b^2
b^2 = 9/4 - 4
b^2 = 9/4 - 16/4
b^2 = -7/4

Since the value of b^2 is negative, we know the hyperbola is vertically oriented.

Now, we can plug the values of the center (-2, 3), a = 2, and b^2 = -7/4 into the standard form equation for a hyperbola, and it becomes:
((x - (-2))^2)/2^2 - ((y - 3)^2)/(-7/4) = 1

Simplifying the equation, we get:
(x + 2)^2/4 - 4(y - 3)^2/7 = 1

Therefore, the equation of the hyperbola is: (x + 2)^2/4 - 4(y - 3)^2/7 = 1.