At which point does the line 2x + sqrt{6}y = | vedclass

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(A) The equation of the line is $2x + \sqrt{6}y = 2$,which can be written as $x = \frac{2 - \sqrt{6}y}{2}$.Substituting this into the hyperbola equati

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$(4, -\sqrt{6})$

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(A) The equation of the line is $2x + \sqrt{6}y = 2$,which can be written as $x = \frac{2 - \sqrt{6}y}{2}$.Substituting this into the hyperbola equation $x^2 - 2y^2 = 4$:$(\frac{2 - \sqrt{6}y}{2})^2 - 2y^2 = 4$$\frac{4 + 6y^2 - 4\sqrt{6}y}{4} - 2y^2 = 4$$4 + 6y^2 - 4\sqrt{6}y - 8y^2 = 16$$-2y^2 - 4\sqrt{6}y - 12 = 0$$y^2 + 2\sqrt{6}y + 6 = 0$$(y + \sqrt{6})^2 = 0$$y = -\sqrt{6}$.Substituting $y = -\sqrt{6}$ into the line equation:$2x + \sqrt{6}(-\sqrt{6}) = 2$$2x - 6 = 2$$2x = 8$$x = 4$.Thus,the point of contact is $(4, -\sqrt{6})$.

At which point does the line $2x + \sqrt{6}y = 2$ touch the hyperbola $x^2 - 2y^2 = 4$?

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