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(A) The given hyperbola is $5 x^2-y^2=5$,which can be rewritten as $\frac{x^2}{1}-\frac{y^2}{5}=1$.Here,$a^2=1$ and $b^2=5$.The equation of a tangent

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$3 x-y+2=0$

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(A) The given hyperbola is $5 x^2-y^2=5$,which can be rewritten 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 $m$ 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 $(2,8)$,we have $8=2 m \pm \sqrt{m^2-5}$,or $(8-2 m)^2 = m^2-5$.Expanding this,$64+4 m^2-32 m = m^2-5$,which simplifies to $3 m^2-32 m+69=0$.Factoring the quadratic equation,$(3 m-23)(m-3)=0$,so $m=3$ or $m=\frac{23}{3}$.For $m=3$,the tangent equation is $y=3 x \pm \sqrt{3^2-5} \Rightarrow y=3 x \pm 2$.Thus,$3 x-y+2=0$ or $3 x-y-2=0$ are the tangents.

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

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