Question

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

Answer 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