A curve, C, passes through the point with coordinates (1, 6) - AQA - A-Level Maths Pure - Question 11 - 2021 - Paper 1

Next, integrate both sides:

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

This can be expressed as:

$-\frac{1}{xy} = \frac{1}{6}x + C$

Now, substitute the point (1, 6) into the integrated equation to determine the constant C:

$-\frac{1}{(1)(6)} = \frac{1}{6}(1) + C$

This simplifies to:

$-\frac{1}{6} = \frac{1}{6} + C$

Thus, by solving, we find:

$C = -\frac{2}{6} = -\frac{1}{3}$

Substituting back the value of C into our integral equation gives:

$-\frac{1}{xy} = \frac{1}{6}x - \frac{1}{3}$

Rearranging this results in:

$y = -\frac{6}{x - 2}$

To find the x-intercept, set y = 0:

$0 = -\frac{6}{x - 2}$

This leads to a contradiction, indicating that C does not intersect the x-axis.

To find the y-intercept, set x = 0:

$y = -\frac{6}{0 - 2} = 3$

Thus, C intersects the y-axis at the point (0, 3).

From the above calculations, we observe that C intersects the coordinate axes at exactly one point, specifically the y-axis at (0, 3). There is no point where C intersects the x-axis.