The equation of the transverse axis of the hy | vedclass

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(B) The given equation is $(x-3)^2+(y+1)^2=(4x+3y)^2$. We can rewrite this as $\sqrt{(x-3)^2+(y+1)^2} = 5 \left| \frac{4x+3y}{5} \right|$. This is of

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$3x-4y=13$

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(B) The given equation is $(x-3)^2+(y+1)^2=(4x+3y)^2$. We can rewrite this as $\sqrt{(x-3)^2+(y+1)^2} = 5 \left| \frac{4x+3y}{5} \right|$. This is of the form $SP = ePM$,where $S(3, -1)$ is the focus,$e = 5$ is the eccentricity,and $4x+3y=0$ is the directrix. The transverse axis is a line passing through the focus $(3, -1)$ and is perpendicular to the directrix $4x+3y=0$. The slope of the directrix is $m_1 = -4/3$. Since the transverse axis is perpendicular to the directrix,its slope is $m_2 = 3/4$. The equation of the transverse axis is $y - (-1) = \frac{3}{4}(x - 3)$. $4(y+1) = 3(x-3) \implies 4y+4 = 3x-9 \implies 3x-4y = 13$. Thus,option $B$ is correct.

The equation of the transverse axis of the hyperbola $(x-3)^2+(y+1)^2=(4x+3y)^2$ is

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