Find all the aspects of hyperbola `16x^(2)-3y^(2)-32x+12y-44=0.`

Answer

`e=sqrt(19//3),"Vertices"-=(1pmsqrt3,2),"Foci"-=(1pmsqrt(19),2), "Directrix:"x-1=pm3//sqrt(19)`

Explanation

We have,
`16(x^(2)-2x)-3(y^(2)-4y)=44`
`"or "16(x-1)^(2)-3(y-2)^(2)=48`
`"or "((x-1)^(2))/(3)-((y-2)^(2))/(16)=1`
Clearly, transverse axis is horizontal, having equation `y-2=0.`
Conjugate axis is `x-1=0 or x=1.`
Centre is (1,2).
Here, `a=sqrt3` and b = 4.
Thus, length of transverse axis is `2sqrt3` and that of conjugate axis is 8.
`"Eccentricity, e"=sqrt(1+(16)/(3))=sqrt((19)/(3))`
Vertices lie at distance a from centre on transverse axis.
Therefore, vertices are `(1pm sqrt3, 2)`
Foci lie at distance ae from centre on transverse axis.
Therefore, foci are `(1pm sqrt(19),2)`.
Equation of directrices are
`x-1= pm(1)/(e)`
`"or "x-1=pm(3)/(sqrt(19)`