Answer:
[tex]\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1[/tex]
Step-by-step explanation:
The standard equation of a horizontal hyperbola with center (h,k) is
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]
The given hyperbola has vertices at (–10, 6) and (4, 6).
The length of its major axis is [tex]2a=|4--10|[/tex].
[tex]\implies 2a=|14|[/tex]
[tex]\implies 2a=14[/tex]
[tex]\implies a=7[/tex]
The center is the midpoint of the vertices (–10, 6) and (4, 6).
The center is [tex](\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)[/tex]
We need to use the relation [tex]a^2+b^2=c^2[/tex] to find [tex]b^2[/tex].
The c-value is the distance from the center (-3,6) to one of the foci (6,6)
[tex]c=|6--3|=9[/tex]
[tex]\implies 7^2+b^2=9^2[/tex]
[tex]\implies b^2=9^2-7^2[/tex]
[tex]\implies b^2=81-49[/tex]
[tex]\implies b^2=32[/tex]
We substitute these values into the standard equation of the hyperbola to obtain:
[tex]\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1[/tex]
[tex]\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1[/tex]
