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`
Text Solution
Verified by Experts
The correct Answer is:
`3x-4y=13`
`(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`
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