At what point/s is the hyperbola \( x y=12 \) has a tangent line parallel to the line \( 3 x-y=0 \) ?

To find the points on the hyperbola \( xy = 12 \) where the tangent line is parallel to the line \( 3x - y = 0 \), we first determine the slope of the given line. The equation can be rewritten in slope-intercept form: \[ y = 3x \] From this, we see that the slope of the line is \( m = 3 \). Next, we need to find the points on the hyperbola where the tangent line has the same slope. We start by implicitly differentiating the equation of the hyperbola \( xy = 12 \). Differentiating both sides with respect to \( x \): \[ \frac{d}{dx}(xy) = \frac{d}{dx}(12) \] Using the product rule on the left side, we have: \[ x \frac{dy}{dx} + y = 0 \] Solving for \( \frac{dy}{dx} \): \[ x \frac{dy}{dx} = -y \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{y}{x} \] We want this derivative to equal the slope of the line, which is 3: \[ -\frac{y}{x} = 3 \] Rearranging gives: \[ y = -3x \] Now, we substitute \( y = -3x \) into the hyperbola equation \( xy = 12 \): \[ x(-3x) = 12 \] This simplifies to: \[ -3x^2 = 12 \quad \Rightarrow \quad x^2 = -4 \] Since \( x^2 = -4 \) has no real solutions, 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 final answer is: \[ \text{There are no points on the hyperbola where the tangent line is parallel to } 3x - y = 0. \]