find the equation of the hyperbola with vertices at (0,±6) and e=5/3 find its foci​

1 thought on “find the equation of the hyperbola with vertices at (0,±6) and e=5/3 find its foci​”

1. Answer:

   vertices of hyperbola=(0,+_6)

   we know vertices of hyperbola =(0,+_ b)

   b=6

   eccentricity e=5/3

   we know

   b^2(e^2-1)=a^2

   6^2((5/3)^2)-1)=a^2

   36(25/9-1)=a^2

   36×16/9=a^2

   a=6×4/3

   a=8

   now

   -x^2/a^2+y^2/b^2=1

   -x^2/8^2+y^2/6^2=1

   -x^2/64+y^2/36=1

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