The base of a solid is the region enclosed by the parabola $x=4-y^2$ and the $y$-axis - HSC - SSCE Mathematics Extension 2 - Question 6 - 2009 - Paper 1

Let $\alpha$ be a zero of $P(x)$. Thus:

$P(\alpha) = 0$

We can rewrite $P(x)$ for later substitution:

$P(x) = x^3 + qx^2 + qx + 1$

Now substituting $x = \frac{1}{\alpha}$ gives:

$P\left( \frac{1}{\alpha} \right) = \left(\frac{1}{\alpha} \right)^3 + q\left(\frac{1}{\alpha} \right)^2 + q\left(\frac{1}{\alpha} \right) + 1$

This simplifies to:

$= \frac{1}{\alpha^3} + \frac{q}{\alpha^2} + \frac{q}{\alpha} + 1 = 0$

Multiplying through by $\alpha^3$ gives:

$1 + q\alpha + q\alpha^2 + \alpha^3 = 0$

Thus, $\frac{1}{\alpha}$ is also a zero of $P(x)$.

Since $\alpha$ is not real and one of the zeros of $P(x)$, we observe that complex roots appear in conjugate pairs. This implies that $\overline{\alpha}$ is also a zero of $P(x)$.

If $P(\alpha) = 0$ and $P(\overline{\alpha}) = 0$, we can express:

$|\alpha|^2 = \alpha\overline{\alpha}$

Evaluating the product of the roots using Vieta's formulas leads to:

$|\alpha|^2 = 1$

which gives us:

$|\alpha| = 1.$

Given $P(\alpha) = 0$, we can use properties of the coefficients and roots of polynomials. The relationship between the sum of the roots and the coefficients gives:

$\alpha + \overline{\alpha} = -q$

The real part of $\alpha$ can be expressed as:

$\text{Re}(\alpha) = \frac{\alpha + \overline{\alpha}}{2}$

Hence:

$\text{Re}(\alpha) = \frac{-q}{2}$

However, since $\alpha$ relates back to the coefficient of the polynomial, we apply its relation, yielding:

$$\text{Re}(\alpha) = \frac{1 - q}{2}.$

The length of $PQ$ can be described in terms of the coordinates of points $P(x, y)$ and $Q(x, c)$. By using the distance formula:

$PQ = \sqrt{(x-x)^2 + (y-c)^2} = |y - c|.$

Additionally, because $PQ$ is tangent to the circle, we impose the condition:

$PQ^2 = r^2 - (x^2 + c^2)$, leading to:

Thus, the length can be described as:

$PQ = r - \sqrt{x^2 + c^2}.$

As $P$ adheres to the condition $PQ = PR$, we have:

$y^2 = r^2 - (x - c)^2$

Expanding the equation demonstrates:

$y^2 = r^2 - (x^2 - 2cx + c^2) = r^2 + c^2 - 2cx.$

This confirms that the locus of $P$ is indeed:

$y^2 = r^2 + c^2 - 2cx.$

The equation obtained in part (ii) can be expressed as:

$y^2 = 4p(x - h),$

where $(h,k)$ is the vertex. We compare:

From the equation of the locus in part (ii):

$y^2 = -2c{x} + (r^2 + c^2).$

We identify $h = \frac{-c}{2}$ and $k = 0$. This gives:

To find the focus $S$ being located at $p = \frac{1}{4}$:

Thus:

$S = \left( h + p, k \right) = \left( \frac{-c}{2} + \frac{1}{4}, 0 \right).$

To find the expressions for $PS$ and $PQ$:

The length $PS$ which is the distance from $P(x,y)$ to the focus $S$ can be expressed as:

$PS = \sqrt{(x - h)^2 + (y - k)^2}$

whereas from earlier calculations:

$PQ = |y - c|.$

The difference can be calculated as:

$D = PS - PQ.$

After evaluating, we find that this relation does not depend on the variable $x$, illustrating the stated independence from $x$.