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 Mechanics - Question 11 - 2021 - Paper 1

To determine where the curve intersects the coordinate axes, we need to analyze the gradient equation given:

$\frac{dy}{dx} = \frac{1}{6}(xy)^2$

1. Separate Variables: Rearrange the gradient equation to separate the variables: $\frac{dy}{(xy)^2} = \frac{1}{6} dx$

2. Integrate Both Sides: Integrating the left side: $\int \frac{dy}{(xy)^2} = \int \frac{1}{6} dx$ This can be simplified to: $\int \frac{1}{y^2} dy = \frac{1}{6} x + C$

3. Find the Integral: The left side integrates to: $-\frac{1}{y} = \frac{1}{6} x + C$

4. Determine the Constant C Using the Point (1, 6): Substitute the point (1, 6) into the integrated equation: $-\frac{1}{6} = \frac{1}{6}(1) + C$ This gives: $C = -\frac{1}{6} - \frac{1}{6} = -\frac{2}{6} = -\frac{1}{3}$

5. Final Equation of the Curve: Substitute back to obtain the equation of C: $y = -\frac{1}{\frac{1}{6}x - \frac{1}{3}}$ Simplifying this leads to the intersection points with the axes.

6. Evaluate the Y-intercept: To find the y-intercept, set (x = 0): $y = -\frac{1}{-\frac{1}{3}} = 3$ Thus, the curve intersects the y-axis at (0, 3).

7. Evaluate the X-intercept: Set (y = 0): From the equation, we see that the curve cannot equal 0 for x because: (x = -\frac{9}{y^2} \Rightarrow \text{undefined when } y = 0) Hence, there is no intersection with the x-axis. This means C only intersects the axes at exactly one point, (0, 3).

8. Conclusion: The coordinates of the intersection with the axes is (0, 3), and the curve does not intersect the x-axis.