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(B) The given hyperbola is $5 x^2-y^2=5$,which can be rewritten in standard form as $\frac{x^2}{1}-\frac{y^2}{5}=1$.Here,$a^2=1$ and $b^2=5$.The equat

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$3 x-y+2=0$

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(B) The given hyperbola is $5 x^2-y^2=5$,which can be rewritten in standard form as $\frac{x^2}{1}-\frac{y^2}{5}=1$.Here,$a^2=1$ and $b^2=5$.The equation of a tangent with slope $m$ to the hyperbola is $y=m x \pm \sqrt{a^2 m^2-b^2}$,which becomes $y=m x \pm \sqrt{m^2-5}$.Since the tangent passes through the point $(2,8)$,we substitute these coordinates into the equation:$8=2 m \pm \sqrt{m^2-5}$$(8-2 m)^2 = m^2-5$$64+4 m^2-32 m = m^2-5$$3 m^2-32 m+69 = 0$$(3 m-23)(m-3) = 0$Thus,$m=3$ or $m=\frac{23}{3}$.For $m=3$,the equation of the tangent is $y=3 x \pm \sqrt{3^2-5} \Rightarrow y=3 x \pm 2$.This gives $3 x-y+2=0$ or $3 x-y-2=0$.Comparing with the options,$3 x-y+2=0$ is the correct equation.

Equation of one of the tangents passing through $(2,8)$ to the hyperbola $5 x^2-y^2=5$ is

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