If a tangent to the hyperbola x^2 - frac{y^2} | vedclass

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(B) The equation of a tangent to the parabola $y^2 = 8x$ with slope $m$ is $y = mx + \frac{2}{m}$.The equation of a tangent to the hyperbola $x^2 - \f

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$y - 2x - 1 = 0$

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(B) The equation of a tangent to the parabola $y^2 = 8x$ with slope $m$ is $y = mx + \frac{2}{m}$.The equation of a tangent to the hyperbola $x^2 - \frac{y^2}{3} = 1$ with slope $m$ is $y = mx \pm \sqrt{m^2 - 3}$.For the lines to be identical,we equate the intercepts: $\frac{2}{m} = \pm \sqrt{m^2 - 3}$.Squaring both sides,we get $\frac{4}{m^2} = m^2 - 3$,which implies $m^4 - 3m^2 - 4 = 0$.Let $t = m^2$,then $t^2 - 3t - 4 = 0$,so $(t - 4)(t + 1) = 0$.Since $m^2 = t$,we have $m^2 = 4$,so $m = \pm 2$.For a positive slope,$m = 2$.Substituting $m = 2$ into the tangent equation $y = mx + \frac{2}{m}$,we get $y = 2x + \frac{2}{2}$,which simplifies to $y = 2x + 1$ or $y - 2x - 1 = 0$.

If a tangent to the hyperbola $x^2 - \frac{y^2}{3} = 1$ is also a tangent to the parabola $y^2 = 8x$,then the equation of such a tangent with a positive slope is:

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