The line 2x + sqrt{6}y = 2 is a tangent to th | vedclass

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(A) The equation of the hyperbola is $x^2 - 2y^2 = 4$,which can be written as $\frac{x^2}{4} - \frac{y^2}{2} = 1$. Comparing this with $\frac{x^2}{a^2

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

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(A) The equation of the hyperbola is $x^2 - 2y^2 = 4$,which can be written as $\frac{x^2}{4} - \frac{y^2}{2} = 1$. Comparing this with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 4$ and $b^2 = 2$. The equation of the tangent at point $(x_1, y_1)$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$. Substituting the values,we get $\frac{xx_1}{4} - \frac{yy_1}{2} = 1$. The given line is $2x + \sqrt{6}y = 2$. Dividing by $2$,we get $x + \frac{\sqrt{6}}{2}y = 1$. Comparing the two equations: $\frac{x_1}{4} = 1 \implies x_1 = 4$. $-\frac{y_1}{2} = \frac{\sqrt{6}}{2} \implies y_1 = -\sqrt{6}$. Thus,the point of contact is $(4, -\sqrt{6})$.

The line $2x + \sqrt{6}y = 2$ is a tangent to the curve $x^2 - 2y^2 = 4$. The point of contact is

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