[Cambridge/2003]
Sketch the curve with parametric equations
$x = 3t, \quad y = \frac{3}{t}.$
The point $P$ on the curve has parameter $t = 2$. The normal at $P$ meets the curve again at the point $Q$.

1. Calculate the coordinates of point $P$ by substituting $t=2$.

$x = 3(2) = 6$
$y = \frac{3}{2}$
So, $P$ is $\left(6, \frac{3}{2}\right)$.

2. Find the derivative $\frac{dy}{dx}$ using the parametric equations.

$\frac{dx}{dt} = 3$
$\frac{dy}{dt} = -\frac{3}{t^2}$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-\frac{3}{t^2}}{3} = -\frac{1}{t^2}$

3. Evaluate the gradient of the curve and use it to find the gradient of the normal at $t=2$.

Gradient of tangent at $t=2$:
$m_t = -\frac{1}{2^2} = -\frac{1}{4}$

Gradient of normal $m_n$ satisfies $m_t m_n = -1$:
$m_n = -\frac{1}{-\frac{1}{4}} = 4$

4. Form the equation of the normal using point $P$ and the normal gradient.

Using the point-slope form $y - y_1 = m_n(x - x_1)$:
$y - \frac{3}{2} = 4(x - 6)$
Multiply the entire equation by 2:
$2y - 3 = 8(x - 6)$
$2y - 3 = 8x - 48$
$2y = 8x - 45 \text{ (Shown)}$

1. Substitute the parametric equations of the curve into the equation of the normal.

Substitute $x = 3t$ and $y = \frac{3}{t}$ into the normal equation $2y = 8x - 45$:
$2\left(\frac{3}{t}\right) = 8(3t) - 45$

2. Rearrange the equation into a standard quadratic form.

$\frac{6}{t} = 24t - 45$
Multiplying the entire equation by $t$ (since $t \neq 0$) and rearranging terms:
$24t^2 - 45t - 6 = 0$
Dividing by 3 to simplify:
$8t^2 - 15t - 2 = 0$

3. Solve the quadratic equation to find the value of $t$ at $Q$.

Factorising the quadratic equation:
$(8t + 1)(t - 2) = 0$
This gives $t = -\frac{1}{8}$ or $t = 2$.

Since $t = 2$ corresponds to the given point $P$, the value of $t$ at the intersection point $Q$ is $-\frac{1}{8}$.