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Endpoints (-5,-4) & (-5,6)
foci (-5,-5) & (-5,7)
Solution
(y-1)^2/25 -(x+5)^2/11 = 1
(y-k)^2/a^2 -(x-h)^2/b^2 = 1 is the standard equation for a hyperbola whose transverse axis is parallel to the y-axis, where (h,k) is the center, the midpoint between the endpoints (vertices) of the hyperbola. k=(6-4)/2= 1
(h,k) = (-5,1)
a=5
b=sqr11
b=sqr(c^2-a^2)
c=-6
the foci are (h,k+c) and (h,k-c) = (-5,-5) and (-5,7)
k+c = 1+c =-5
c =-5-1=-6
b^2 =36-25=11
b=sqr11
You could also try to get the equation by using the definition of a hyperbola as the locus of all points such that their difference in distance from the 2 foci are equal. In this case, take either vertex and calculate the distance to each focus, 1 and 11. Their difference = 10. The hyperbola is all (x,y) points such that their distance from one focus minus the distance to the other =10.