Find the coordinates of the foci and the center of the hyperbola

Explanation

We have
`((3x-4y-12)^(2))/(100)-((4x+3y-12)^(2))/(225)=1`
`"or "((3x-4y-12)/sqrt(3^(2)+(-4)^(2)))^(2)/(4)-((4x+3y-12)/(sqrt(4^(2)+3^(2))))^(2)/(9)=1`
Transverse axis is `4x+3y-12=0` and conjugate axis is
`3x-4y-12=0.`
Centre is intersection of axes of hyperbola, which is `((84)/(25),-(12)/(25))`.
Here, a = 2 and b = 3.
`therefore" "e^(2)=1+(b^(2))/(a^(2))=1+(9)/(4)=(13)/(4)`
`therefore" "e=(sqrt(13))/(2)` ltbr? So, `ae=sqrt(13).`
Now, slope of transverse axis is `-(4)/(3)=tan theta`, where `theta` is inclination of line with x-axis.
Foci lie on transverse axis at distance 'ae' from the centre.
Therefore, foci are `((84)/(25)pmsqrt(13)cos theta,(-12)/(25)pmsqrt(13)sintheta)`
`-=((84)/(25)pmsqrt(13)xx(3)/(5),(-12)/(25)pmsqrt(13)xx(4)/(5))`
`-=((84pm15sqrt(13))/(25),(-12pm20sqrt(13))/(25))`
Directrix lie at distance `(a)/(e)` from the conjugate axis.
Now, `(a)/(e)=(2)/(sqrt(13)//2)=(4)/(sqrt(13)).`
Thus, distnace between conjugate axis and directrix is `(4)/(sqrt(13)).`
So, equations of directrices are given by `3x-4y+c=0,` where
`(|c+12|)/(sqrt(3^(2)+(-4)^(2)))=(4)/(sqrt(13)).`
`therefore" "c=-12pm(20)/(sqrt(13))`
So, equations of directrices are `3x-4y-12pm(20)/(sqrt(13))=0.`