A curve, C, passes through the point with coordinates (1, 6) The gradient of C is given by dy/dx = (1/6)(xy)^2 Show that C intersects the coordinate axes at exactly one point and state the coordinates of this point - AQA - A-Level Maths Pure - Question 11 - 2021 - Paper 1

To show that the curve intersects the coordinate axes, we need to find the points where the curve meets the x-axis and y-axis.

1. Finding the equation of the curve: We start with the gradient of the curve: $\frac{dy}{dx} = \frac{1}{6}(xy)^2$ We can separate the variables and integrate:

   $\int \frac{dy}{(y)^2} = \int \frac{1}{6} x dx$

   This leads to:

   $-\frac{1}{y} = \frac{1}{6} \frac{x^2}{2} + C$

   Rearranging gives:

   $y = -\frac{1}{\frac{x^2}{12} + C}$

2. Substituting the point (1, 6): To find the constant, we substitute the point (1, 6) into the equation:

   $6 = -\frac{1}{\frac{1^2}{12} + C}$

   This leads to:

   $\frac{1}{6} = \frac{1}{\frac{1}{12} + C}\implies \frac{1}{12} + C = 6\implies C = 6 - \frac{1}{12} = \frac{72}{12} - \frac{1}{12} = \frac{71}{12}$

3. Conclusion regarding the x-intercept: We now determine when the curve intersects the x-axis, which occurs when y = 0:

   Setting the equation for y to zero does not yield a valid solution because as ( y ) cannot equal zero in the re-arranged equation. Thus, the curve does not intersect the x-axis (i.e., no real solution for x results).

4. Finding the y-intercept: We find the y-intercept by setting x = 0:

   $y = -\frac{1}{\frac{0^2}{12} + \frac{71}{12}} = -\frac{1}{\frac{71}{12}} = -\frac{12}{71}$

   Thus, the curve intersects the y-axis at (0, -12/71).

5. Final Evaluation: The curve intersects the coordinate axes only at the y-axis at one point (0, -12/71) and not at the x-axis.