If y=x+sqrt{2} is a tangent to the hyperbola | vedclass

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(B) The given hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{2} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we hav

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$x= \pm \sqrt{\frac{8}{3}}$

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(B) The given hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{2} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $b^2 = 2$. The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$. Given the tangent line $y = x + \sqrt{2}$,we have $m = 1$ and $c = \sqrt{2}$. Substituting these values into the condition: $(\sqrt{2})^2 = a^2(1)^2 - 2$ $\Rightarrow 2 = a^2 - 2$ $\Rightarrow a^2 = 4$. The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{2}{4}} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}$. The equations of the directrices are $x = \pm \frac{a}{e}$. Since $a = \sqrt{4} = 2$,we have $x = \pm \frac{2}{\sqrt{3/2}} = \pm 2 \sqrt{\frac{2}{3}} = \pm \sqrt{4 	imes \frac{2}{3}} = \pm \sqrt{\frac{8}{3}}$.

If $y=x+\sqrt{2}$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{2}=1$,then the equations of its directrices are

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