a) Prove by induction that $$n^3 + (n + 1)^3 + (n + 2)^3$$ is divisible by 9 for $n = 1, 2, 3, \ldots$ b) Consider the variable point $P(2at, a t^2)$ on the parabola $x^2 = 4ay$ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2001 - Paper 1

To find the equation of the normal, we first need the slope of the tangent line. For the parabola $x^2 = 4ay$, the derivative (slope) at the point $P(2at, at^2)$ is given by:

$\frac{dy}{dx} = \frac{x}{2a}$

At point $P$, $x = 2at$, so:

$\frac{dy}{dx}|_{P} = \frac{2at}{2a} = t$

The slope of the normal is then $-\frac{1}{t}$. Using point-slope form:

$y - at^2 = -\frac{1}{t}(x - 2at)$

Rearranging gives us:

$x + y = at^2 + 2at,$ as required.

Let $Q(2a \lambda, a \lambda^2)$ on the parabola. The slope of the normal at $Q$:

From the same derivatives, we have:

$\frac{dy}{dx}|_{Q} = \frac{2a \lambda}{2a} = \lambda$

Thus, the slope of its normal is $-\frac{1}{\lambda}$.

Since normals must satisfy:

$\lambda \cdot t = -1,$ hence:

$\lambda = -\frac{1}{t}$

Substituting $\\lambda$ back into the coordinates gives:

$Q(2a(-\frac{1}{t}), a(-\frac{1}{t})^2)$

which simplifies to $Q(-\frac{2a}{t}, \frac{a}{t^2})$.

From part (ii), the normals intersect at $R$ with coordinates:

$x = a\left(t - \frac{1}{t}\right), \, y = a\left(t^2 + 1 + \frac{1}{t^2}\right).$

We can express $y$ in terms of $x$. Substituting $x$ into the $y$ coordinate:

Note that:

$t^2 + 1 + \frac{1}{t^2} = \left(\frac{x}{a} + 1\right).$

Thus, from this substitution we can eliminate $t$ to express $y$ solely in terms of $x$, giving us the Cartesian form of the locus.

To find the Cartesian form of the locus, we express both $x$ and $y$ in terms of $t$:

Starting from:

$x = a\left(t - \frac{1}{t}\right)$ and $y = a\left(t^2 + 1 + \frac{1}{t^2}\right).$

Upon rearranging, we express $t$ from $x$, and subsequently substitute back into $y$:

This leads us to derive an equation involving only $x$ and $y$, yielding the Cartesian form of the locus.