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

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Aicharya · Accepted Answer

$\begin{aligned} (x-3)^2+(y+1)^2 & =(4 x+3 y)^2 \ (x-3)^2+(y+1)^2 & =25\left(\frac{4 x+3 y}{5}\right)^2 \ (x-3)^2+(y+1)^2 & =25\left(\frac{4 x+3 y}{25}\right)^2 \ \sqrt{(x-3)^2+(y+1)^2} & =5\left(\frac{4 x+3 y}{\sqrt{5}}\right)^2 \end{aligned}$ $\therefore$ It is of the form $S P=\mathrm{ePM}$ $\therefore$ Focus $=(3,-1)$ Equation of directrix $=4 x+3 y$ Since, Transverse axis perpendicular to directrix and passing through focus. $\begin{aligned} \therefore \quad 3 x-4 y+k & =0 \ 3(3)-4(-1)+k & =0 \ 13+k & =0 \ k & =-13 \end{aligned}$ $\therefore$ Required Line is $3 x-4 y-13=0$ $3 x-4 y=13$ Hence, option (b) is correct.

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

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