The equation of a tangent to the hyperbola 5x | vedclass

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

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$23x-3y-22=0$

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(A, C) The given equation of the hyperbola is $5x^{2}-y^{2}=5$,which can be written 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 $y=mx \pm \sqrt{a^{2}m^{2}-b^{2}}$,which becomes $y=mx \pm \sqrt{m^{2}-5}$. Since the tangent passes through $(2, 8)$,we have $8=2m \pm \sqrt{m^{2}-5}$. Rearranging gives $8-2m = \pm \sqrt{m^{2}-5}$. Squaring both sides: $(8-2m)^{2} = m^{2}-5$. $64+4m^{2}-32m = m^{2}-5$. $3m^{2}-32m+69=0$. Solving for $m$: $(3m-23)(m-3)=0$,so $m=3$ or $m=\frac{23}{3}$. For $m=3$,the tangent equation is $y-8=3(x-2)$ $\Rightarrow y-8=3x-6$ $\Rightarrow 3x-y+2=0$. For $m=\frac{23}{3}$,the tangent equation is $y-8=\frac{23}{3}(x-2)$ $\Rightarrow 3y-24=23x-46$ $\Rightarrow 23x-3y-22=0$. Both $3x-y+2=0$ and $23x-3y-22=0$ are valid tangents.

The equation of a tangent to the hyperbola $5x^{2}-y^{2}=5$ which passes through the external point $(2, 8)$ is:

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