To find the points where the gradient of the tangent is 8, we first need to find the derivative of \(y = \frac{x^2}{1 - 3x}\).
Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = x^2\) and \(v = 1 - 3x\).
\(\frac{du}{dx} = 2x\) and \(\frac{dv}{dx} = -3\).
Thus, the derivative \(\frac{dy}{dx} = \frac{(1 - 3x)(2x) - x^2(-3)}{(1 - 3x)^2} = \frac{2x - 6x^2 + 3x^2}{(1 - 3x)^2} = \frac{2x - 3x^2}{(1 - 3x)^2}\).
Set \(\frac{2x - 3x^2}{(1 - 3x)^2} = 8\) and solve for \(x\).
\(2x - 3x^2 = 8(1 - 3x)^2\)
\(2x - 3x^2 = 8(1 - 6x + 9x^2)\)
\(2x - 3x^2 = 8 - 48x + 72x^2\)
\(75x^2 - 50x + 8 = 0\)
Factor the quadratic: \((15x - 4)(5x - 2) = 0\)
Solutions for \(x\) are \(x = \frac{2}{5}\) and \(x = \frac{4}{15}\).
Substitute back to find \(y\):
For \(x = \frac{2}{5}\), \(y = \frac{(\frac{2}{5})^2}{1 - 3(\frac{2}{5})} = \frac{\frac{4}{25}}{1 - \frac{6}{5}} = \frac{\frac{4}{25}}{-\frac{1}{5}} = -\frac{4}{5}\).
For \(x = \frac{4}{15}\), \(y = \frac{(\frac{4}{15})^2}{1 - 3(\frac{4}{15})} = \frac{\frac{16}{225}}{1 - \frac{12}{15}} = \frac{\frac{16}{225}}{\frac{1}{5}} = \frac{16}{45}\).
Thus, the points are \(\left( \frac{2}{5}, -\frac{4}{5} \right)\) and \(\left( \frac{4}{15}, \frac{16}{45} \right)\).
