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

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Question

(A) The equation of the curve is $x^2 - 2y^2 = 4$. Differentiating with respect to $x$,we get $2x - 4y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx}

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

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(A) The equation of the curve is $x^2 - 2y^2 = 4$. Differentiating with respect to $x$,we get $2x - 4y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = \frac{x}{2y}$. The slope of the tangent line $2x + \sqrt{6}y = 2$ is $-\frac{2}{\sqrt{6}} = -\frac{\sqrt{6}}{3}$. Let the point of contact be $(x_1, y_1)$. Then $\frac{x_1}{2y_1} = -\frac{\sqrt{6}}{3}$,so $3x_1 = -2\sqrt{6}y_1$,or $y_1 = -\frac{3x_1}{2\sqrt{6}} = -\frac{\sqrt{6}x_1}{4}$. Since the point $(x_1, y_1)$ lies on the curve,$x_1^2 - 2(-\frac{\sqrt{6}x_1}{4})^2 = 4$. $x_1^2 - 2(\frac{6x_1^2}{16}) = 4 \implies x_1^2 - \frac{3}{4}x_1^2 = 4 \implies \frac{1}{4}x_1^2 = 4 \implies x_1^2 = 16$. Thus $x_1 = 4$ or $x_1 = -4$. If $x_1 = 4$,then $y_1 = -\frac{\sqrt{6}(4)}{4} = -\sqrt{6}$. Checking the line equation: $2(4) + \sqrt{6}(-\sqrt{6}) = 8 - 6 = 2$. This satisfies the equation. 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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