Show that the tangent to the curve \( y=\frac{2}{3 x-1} \) at the point \( P(1,1) \) meets the curve only at one point.

To find the tangent to the curve \( y = \frac{2}{3x - 1} \) at the point \( P(1, 1) \), we first calculate the derivative to find the slope at that point. Using the quotient rule, we differentiate: \[ y' = \frac{(0)(3x - 1) - (2)(3)}{(3x - 1)^2} = \frac{-6}{(3x - 1)^2}. \] Now, evaluating the derivative at \( x = 1 \): \[ y'(1) = \frac{-6}{(3(1) - 1)^2} = \frac{-6}{(3 - 1)^2} = \frac{-6}{4} = -\frac{3}{2}. \] The equation of the tangent line at point \( P(1, 1) \) is: \[ y - 1 = -\frac{3}{2}(x - 1). \] Simplifying yields: \[ y = -\frac{3}{2}x + \frac{5}{2}. \] To show that this tangent line meets the curve only at one point, we set it equal to the original curve: \[ -\frac{3}{2}x + \frac{5}{2} = \frac{2}{3x - 1}. \] Multiplying both sides by \( 3x - 1 \) to eliminate the fraction gives: \[ (-\frac{3}{2}x + \frac{5}{2})(3x - 1) = 2. \] Expanding the left side results in: \[ -\frac{9}{2}x^2 + \frac{15}{2}x + \frac{3}{2}x - \frac{5}{2} = 2, \] which simplifies to: \[ -\frac{9}{2}x^2 + \frac{12}{2}x - \frac{5}{2} = 2. \] Multiplying through by 2 to clear the fraction: \[ -9x^2 + 12x - 5 = 4 \rightarrow -9x^2 + 12x - 9 = 0. \] This can be simplified to: \[ 9x^2 - 12x + 9 = 0. \] Calculating the discriminant \( D \): \[ D = b^2 - 4ac = (-12)^2 - 4(9)(9) = 144 - 324 = -180. \] Since the discriminant is negative, this implies that there are no real solutions, meaning the tangent line meets the curve only at the point \( P(1, 1) \). Thus, we have shown that the tangent meets the curve only at a single point.