If the line 2x + sqrt{6}y = 2 touches the hyp | vedclass

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(B) The given line is $2x + \sqrt{6}y = 2$,which can be written as $2x + \sqrt{6}y - 2 = 0$.The equation of the hyperbola is $x^2 - 2y^2 = 4$.Let the

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

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(B) The given line is $2x + \sqrt{6}y = 2$,which can be written as $2x + \sqrt{6}y - 2 = 0$.The equation of the hyperbola is $x^2 - 2y^2 = 4$.Let the point of contact be $(x_1, y_1)$.The equation of the tangent to the hyperbola $x^2 - 2y^2 = 4$ at $(x_1, y_1)$ is given by $xx_1 - 2yy_1 = 4$,or $xx_1 - 2yy_1 - 4 = 0$.Since this line is the same as the given line $2x + \sqrt{6}y - 2 = 0$,the coefficients must be proportional:$\frac{x_1}{2} = \frac{-2y_1}{\sqrt{6}} = \frac{-4}{-2}$$\frac{x_1}{2} = 2 \Rightarrow x_1 = 4$$\frac{-2y_1}{\sqrt{6}} = 2$ $\Rightarrow -2y_1 = 2\sqrt{6}$ $\Rightarrow y_1 = -\sqrt{6}$Thus,the point of contact is $(4, -\sqrt{6})$.

If the line $2x + \sqrt{6}y = 2$ touches the hyperbola $x^2 - 2y^2 = 4$,then the coordinates of the point of contact are

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