Consider the hyperbola $H$ with equation \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \) - HSC - SSCE Mathematics Extension 2 - Question 3 - 2001 - Paper 1

The equations of the directrices for a hyperbola are given by: $x = \pm \frac{a}{e}$ Substituting our values: $x = \pm \frac{3}{\frac{5}{3}} = \pm \frac{9}{5}$

Thus, the equations of the directrices are: $x = \frac{9}{5} \quad \text{and} \quad x = -\frac{9}{5}$

The asymptotes are given by: $y = \pm \frac{b}{a} x$ Thus: $y = \pm \frac{4}{3} x$

A sketch of the hyperbola would show it opening left and right, with the center at the origin $(0,0)$, intersecting the x-axis at $(3,0)$ and $(-3,0)$, with foci at $(5,0)$ and $(-5,0)$, directrices at $x = \frac{9}{5}$ and $x = -\frac{9}{5}$, and asymptotes forming lines at $y = \frac{4}{3}x$ and $y = -\frac{4}{3}x$.

Using the identities: $\alpha + \beta + \gamma = 3$ $\alpha^2 + \beta^2 + \gamma^2 = 1$ We can express: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{1}{2}((\alpha + \beta + \gamma)^2 - (\alpha^2 + \beta^2 + \gamma^2))$ Substituting the known values gives us: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{1}{2}(3^2 - 1) = \frac{1}{2}(9 - 1) = \frac{8}{2} = 4$

Given the polynomial $x^3 - 3x^2 + 4x - 2 = 0$, we find that the roots are determined by Vieta's formulas:

• The sum of the roots $\alpha + \beta + \gamma = 3$,
• The sum of the products of the roots taken two at a time $\alpha\beta + \beta\gamma + \gamma\alpha = 4$,
• The product of the roots $\alpha\beta\gamma = 2$ (also derived from rearranging the original equations). This verifies that $\alpha, \beta$, and $\gamma$ meet the cubic equation's root conditions.

To find the actual values of $\alpha, \beta$, and $\gamma$, we can substitute into the cubic equation: $x^3 - 3x^2 + 4x - 2 = 0$ By testing for rational roots, it can be solved using synthetic division or numerical methods. The solutions yield: $\alpha = 1, \beta = 1, \gamma = 1$, which fulfill the equations provided earlier.

Using the method of cylindrical shells, the volume $V$ is given by: $V = 2\pi \int_{0}^{\pi} y \cdot r \, dx$ In this case, we consider the radius $r = y$ and the height of the cylindrical shell. We integrate: $\int_{0}^{\pi} \sin x \, dx = -\cos x \big|_0^{\pi} = -(-1 - 1) = 2$ Thus: $V = 2\pi \cdot 2 = 4\pi$ indicating the volume of the solid of revolution.