The equation of the hyperbola with vertices at `(0, +-6)` and eccentricity `5/3` is….. And its foci are …….

Explanation

Let the equation of the hyperbola be -`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
Then vertices=(0,`pm`b)=(0,`pm`6)
`therefore b=6 and e=5//3`
`becausee=sqrt(1+(a^(2))/(b^(2)))rArr25/9=1+(a^(2))/36`
`rArr (25-9)/9=(a^(2))/36rArr16=(a^(2))/4rArra^(2)=48`
So, the equation of hyperbola is
`(-x^(2))/48+(y^(2))/36=1rArr(y^(2))/36-(x^(2))/48=1`
`because` Foci=(0,`pm`be)=(0,`pm5/3xx6`)=(0,`pm`10)