Find the coordinates of the foci, the eccentricity, the latus rectum, and the equations of directrices for the hyperbola `9x^2-16 y^2-72 x+96 y-144=0`

Explanation

We have hyperbola `9x^(2)-16y^(2)-72x+96y-144=0.`
`"or "9(x^(2)-8x+16)-16(y^(2)-6y+9)=144`
`"or "((x-4)^(2))/(4^(2))-((y-3)^(2))/(3^(2))=1.`
Cleary, centre of the hyperbola is C(4, 3).
Transverse axis is `y-3=0 or y=3 ` and conjugate axis is `x-4=0 or x=4`.
Also, semi transverse axis is a = 4 and semi conjugate axis is b = 3.
From `b^(2)=a^(2)(e^(2)-1)`, we get eccentricity `e=5//4.`
Vertices are at destance 'a' units from the centre.
So, vertices are `(4-4,3)-=(0,3) and (4+4,3)-=(8,7)`
Foci are at distance 'ae' from the centre.
Now, ae = 5.
So, foci are `(4-5,3)-=(-1,3) and (4+5,3)-=(9,3).`
Two directrices are distance 'a/e' from the conjugate axis.
Thus, directrices are given by
`x-4=pma//e`
`rArr" "x-4=pm16//5.`
`rArr" "5x-36=0 and 5x-4=0`
Length of latus rectum `=2b^(2)//a=2xx9//4=9//2`.