The equation of the transvers axis of the hyperbola `(x-3)^2+(y=1)^2+(4x+3y)^2` is `x+3y=0` (b) `4x+3y=9` `3x-4y=13` (d) `4x+3y=0`

Answer

`3x-4y=13`

Explanation

`(x-3)^(2)+(y+1)^(2)=(4x+3y)^(2)`
`"or "sqrt((x-3)^(2)+(y+1)^(2))=5((|4x+3y|)/(sqrt(4^(2)+3^(2))))`
Clearly, it is hyperbola with focus at (3, -1) and directrix `4x+3y=0`.
Transverse axis is perpendicular to directrix and passes through focus.
So, its equation is
`y+1=(3)/(4)(x-3)`
`"or "3x-4y=13`