Question 1194067
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The vertices of a hyperbola are the points (−3, 2) and (−3, −2) and the length of its conjugate is 6.
Find the equation of the hyperbola, the coordinates of its foci, and its eccentricity.
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<pre>
From the given info, the center of the hyperbola is the point (-3,0)

and its <U>real axis</U> lies on vertical line x = -3.


    Thus the hyperbola is "vertical", opened up and down.  


Real semi-axis length is half the distance between the vertices  a = {{{(2-(-2))/2}}} = {{{4/2}}} = 2.


Also, from the given part, the length of its conjugate axis is 6; hence, conjugate semi-axis length is  b = 6/2 = 3.


Now the standard form equation of the hyperbola is

    {{{y^2/2^2}}} - {{{(x+3)^2/3^2}}} = 1.    <U>ANSWER</U>


                            P L O T


    {{{drawing( 400, 400, -10, 4, -5, 5,
          circle(-3, 0, 0.1), circle(-3, sqrt(13), 0.1), circle(-3, -sqrt(13), 0.1),
          circle(-3, 2, 0.1), circle(-3, -2, 0.1), 

      graph( 400, 400, -10, 4, -5, 5,
          2*sqrt(1+((x+3)^2/9)),
         -2*sqrt(1+((x+3)^2/9)))
)}}}


                 Hyperbola {{{y^2/2^2}}} - {{{(x+3)^2/3^2}}} = 1



In the plot, we can not see the imaginary axis and imaginary semi-axes: they are invisible.  

But we can show the foci.  They are  ({{{-3}}},{{{sqrt(13)}}})  and  ({{{-3}}},{{{-sqrt(13)}}}),  where 13 = {{{a^2}}} + {{{b^2}}} = {{{2^2}}} + {{{3^2}}}.
</pre>

Solved.