Let P(x_0, y_0) be the point on the hyperbola | vedclass

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(C) The equation of the hyperbola is $3x^2 - 4y^2 = 36$,which can be written as $\frac{x^2}{12} - \frac{y^2}{9} = 1$. Here $a^2 = 12$ and $b^2 = 9$,so

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$-9$

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(C) The equation of the hyperbola is $3x^2 - 4y^2 = 36$,which can be written as $\frac{x^2}{12} - \frac{y^2}{9} = 1$. Here $a^2 = 12$ and $b^2 = 9$,so $a = 2\sqrt{3}$ and $b = 3$. The parametric coordinates of a point $P$ on the hyperbola are $(a \sec 	heta, b 	an 	heta) = (2\sqrt{3} \sec 	heta, 3 	an 	heta)$. The slope of the tangent at $P$ must be equal to the slope of the line $3x + 2y = 1$,which is $m = -\frac{3}{2}$. The slope of the tangent at $(x_0, y_0)$ is $\frac{dy}{dx} = \frac{3x_0}{4y_0} = -\frac{3}{2} \implies \frac{x_0}{2y_0} = -1 \implies x_0 = -2y_0$. Substitute $x_0 = -2y_0$ into the hyperbola equation: $3(-2y_0)^2 - 4y_0^2 = 36 \implies 12y_0^2 - 4y_0^2 = 36 \implies 8y_0^2 = 36 \implies y_0^2 = \frac{9}{2} \implies y_0 = \pm \frac{3}{\sqrt{2}}$. Since the line $3x + 2y = 1$ is in the first and fourth quadrants,we check the distance. For $y_0 = -\frac{3}{\sqrt{2}}$,$x_0 = -2(-\frac{3}{\sqrt{2}}) = \frac{6}{\sqrt{2}} = 3\sqrt{2}$. Then $\sqrt{2}(y_0 - x_0) = \sqrt{2}(-\frac{3}{\sqrt{2}} - \frac{6}{\sqrt{2}}) = -3 - 6 = -9$.

Let $P(x_0, y_0)$ be the point on the hyperbola $3x^2 - 4y^2 = 36$ which is nearest to the line $3x + 2y = 1$. Then $\sqrt{2}(y_0 - x_0)$ is equal to:

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