Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).

Accepted Solution

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]