The slope of the tangent drawn from the point | vedclass

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(C) The equation of the hyperbola is $2x^2-y^2=4$,which can be written as $\frac{x^2}{2}-\frac{y^2}{4}=1$. Here,$a^2=2$ and $b^2=4$. The equation of a

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$-1 \pm \sqrt{6}$

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(C) The equation of the hyperbola is $2x^2-y^2=4$,which can be written as $\frac{x^2}{2}-\frac{y^2}{4}=1$. Here,$a^2=2$ and $b^2=4$. The equation of a tangent with slope $m$ to the hyperbola is $y=mx \pm \sqrt{a^2m^2-b^2}$,which becomes $y=mx \pm \sqrt{2m^2-4}$. Since the tangent passes through the point $(1,1)$,we substitute $x=1$ and $y=1$: $1 = m(1) \pm \sqrt{2m^2-4}$ $1-m = \pm \sqrt{2m^2-4}$ Squaring both sides,we get: $(1-m)^2 = 2m^2-4$ $1+m^2-2m = 2m^2-4$ $m^2+2m-5 = 0$ Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $m = \frac{-2 \pm \sqrt{2^2-4(1)(-5)}}{2(1)} = \frac{-2 \pm \sqrt{4+20}}{2} = \frac{-2 \pm \sqrt{24}}{2} = \frac{-2 \pm 2\sqrt{6}}{2} = -1 \pm \sqrt{6}$.

The slope of the tangent drawn from the point $(1,1)$ to the hyperbola $2x^2-y^2=4$ is

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