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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` |
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Answer» Correct Answer - `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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