In each of the find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.
`16x^(2)-9y^(2)=576`

Answer

`e=5//4, "Vertices"-=(0, pm4), "Foci"-=(0, pm5), L.R. = 9//2, "Directrix;y"=pm16//5`

Explanation

We have hyperbola
`16x^(2)-9y^(2)=-144`
`"or "(x^(2))/(9)-(y^(2))/(16)=-1`
This equation is of the form `(x^(2))/(a^(2))-(y^(2))/(16)=-1`.
Hence, x-axis is the conjugate axis and y-axis is the tranverse axis.
Now, `a^(2)=9,b^(2)=16." So, "a=3,b=4.`
Length of transverse axis = 2b = 8
Length of conjugate axis = 2a = 6
Eccentricity, `e=sqrt(1+(a^(2))/(b^(2)))=sqrt(1+(9)/(16))=(5)/(4)`
Vertics are `(0 pmb) or (0, pm4)`.
Foci are `(0, pm be) or (0, pm 5)`
Length of latus rectum `=(2a^(2))/(b)=(2(3)^(2))/(4)=(9)/(2)`
Equation of directrices are
`y=pm(b)/(e)`
`"or "y=pm(4)/((5//4))`
`"or "y=pm(16)/(5)`