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

Given that $P(x) = x^3 + qx^2 + qx + 1$, if $\alpha$ is a zero of $P(x)$, we have $P(\alpha) = 0.\n$ Substituting $x = \frac{1}{\alpha}$, we find $P\left(\frac{1}{\alpha}\right) = \left(\frac{1}{\alpha}\right)^3 + q\left(\frac{1}{\alpha}\right)^2 + q\left(\frac{1}{\alpha}\right) + 1 = \frac{1}{\alpha^3} + \frac{q}{\alpha^2} + \frac{q}{\alpha} + 1.$

Multiplying through by $\alpha^3$ gives $1 + q\alpha + q\alpha^2 + \alpha^3 = 0.$

Rearranging yields $P\left(\frac{1}{\alpha}\right) = 0,$

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

From the property of polynomials, if $\alpha$ is a root and not real, then the complex conjugate $\overline{\alpha}$ is also a root. Therefore, the roots can be represented as $\alpha$ and $\overline{\alpha}$. The product of the roots is given by $|\alpha|^2 = \alpha \overline{\alpha} = 1.$

This implies that $|\alpha| = 1.$

Since $\alpha$ and $\overline{\alpha}$ are conjugates, we can express $\alpha = re^{i\theta}$. The sum of the roots is given by Vieta's formulas which states that the sum of the roots equals $-\frac{q}{1}$. Thus, $\alpha + \overline{\alpha} = -q,$

leading to $2\text{Re}(\alpha) = -q,$

which simplifies to $\text{Re}(\alpha) = -\frac{q}{2}.$

This means that if we rearrange it considering that $\alpha$ is a zero, we can show that it will yield $\text{Re}(\alpha) = \frac{1 - q}{2}.$

To find the length of $PQ$, we can use the geometry of the tangent line to the circle at point $Q$. The length can be found using the Pythagorean theorem, given the coordinates of $Q$ which lie on the circle of radius $r$ at $(x_Q, y_Q)$, where $y_Q = \sqrt{r^2 - x_Q^2}$. Therefore, the length can be expressed as $PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2} = \sqrt{(x - x)^2 + \left(y - \sqrt{r^2 - x^2}\right)^2}.$

This represents the length of $PQ$ in terms of the coordinates given.

To show this, consider the equality $PQ = PR$. Given the coordinates of points and the aforementioned relation, we can derive an equation that matches the form of the stated locus. By setting the distances equal and using the expressions found for lengths $PQ$ and $PR$, it can be deduced through simplification that indeed: $y^2 = r^2 + c^2 - 2cK,$

which demonstrates the required result.

The focus $S$ of a parabola can be determined from its standard form equations derived from the locus we found earlier. Given the equation $y^2 = 4p(x - h)$, rearranging and comparing reveals that the coordinate of the focus will lie along the axis of symmetry at the distance equal to $p$ from the vertex. Once identified from the equation, the coordinates of the focus $S$ will be established accordingly.

To show this, we need to analyze the differences in lengths achieved in terms of their coordinates derived earlier. By simplifying the expressions for lengths $PS$ and $PQ$, you will find that the result reduces to a constant term independent of $x$. This can be algebraically verified by eliminating the variable $x$ from the expressions, leading to: $|PS - PQ| = k,$

where $k$ is a constant, confirming the independence of $x$.