At what point/s is the hyperbola \( x y=12 \) has a tangent line parallel to the line \( 3 x-y=0 \) ? a. \( (-2,6) \) or \( (2,-6) \). b. \( (-2,6) \) or \( (2,6) \). d. \( (-2,-6) \) or \( (2,6) \). e. Nuch point does not exist in real numbers.

To find the points on the hyperbola \( xy = 12 \) where the tangent line is parallel to the line \( 3x - y = 0 \), we first need to determine the slope of the line \( 3x - y = 0 \). Rearranging this equation gives us: \[ y = 3x \] Thus, the slope of the line is \( 3 \). Next, we need to find the slope of the tangent line to the hyperbola \( xy = 12 \). We can do this by implicitly differentiating the equation of the hyperbola with respect to \( x \): \[ \frac{d}{dx}(xy) = \frac{d}{dx}(12) \] Using the product rule on the left side, we get: \[ x \frac{dy}{dx} + y = 0 \] Solving for \( \frac{dy}{dx} \): \[ x \frac{dy}{dx} = -y \implies \frac{dy}{dx} = -\frac{y}{x} \] We want this slope to equal \( 3 \): \[ -\frac{y}{x} = 3 \] This can be rearranged to: \[ y = -3x \] Now we have two equations to work with: 1. The equation of the hyperbola: \( xy = 12 \) 2. The equation of the line: \( y = -3x \) We can substitute \( y = -3x \) into the hyperbola equation: \[ x(-3x) = 12 \] This simplifies to: \[ -3x^2 = 12 \implies x^2 = -4 \] Since \( x^2 = -4 \) has no real solutions (as the square of a real number cannot be negative), we conclude that there are no points on the hyperbola \( xy = 12 \) where the tangent line is parallel to the line \( 3x - y = 0 \). Thus, the answer is: **e. No such point exists in real numbers.**