Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.`5y^2-9x^2=36`

Explanation

Equation of hyperbola
`5y^(2)-9x^(2)=36`
`rArr(5y^(2))/(36)-(9x^(2))/(36)=1rArr(y^(2))/(36//5)-(x^(2))/(4)=1`
`:.` The transverse axis of hyperbola is along y-axis.
Comparing with `(y^(2))/(b^(2))-(x^(2))/(a^(2))=1`
`a^(2)=4," "b^(2)=(36)/(5)`
`rArr" "a=2," "b=(6)/(sqrt(5))`
`:. "Vertices"-=(0,pmb)-=(0,pm(6)/(sqrt(5)))`
Eccentricity `e=sqrt(1+(a^(2))/(b^(2)))=sqrt(1+(4)/(36//5))`
`=sqrt((14)/(9))=(sqrt(14))/(3)`
Now, `be=(6)/(sqrt(5))xx(sqrt(14))/(3)=2sqrt((14)/(5))`
`:. "Coordinates of foci" -=(0,pmbe)-=(0,pm2sqrt((14)/(5)))`
Length of latus rectum `=(2a^(2))/(b)`
`=(2xx4)/(6//sqrt(5))=(4sqrt(5))/(3)`