Let $\, ilde{ ext{w}} \, $ be the complex number satisfying $ \, ext{w}^3 = 1 \, $ and $ \, ext{Im}( ext{w}) > 0 \, $ - HSC - SSCE Mathematics Extension 2 - Question 6 - 2008 - Paper 1

To find the equation of the line $\ell$, we start from the definition of the slope from the point $P(\sec \theta, b \tan \theta)$:

The slope $m$ of the line is computed using the points:

$m = \frac{dy}{dx} = \frac{b \tan \theta - 0}{\sec \theta - (-a)} = \frac{b \tan \theta}{\sec \theta + a}$

Now, applying the point-slope form of the equation of a line, we have:

$y - \left( b \tan \theta \right) = m \left( x - \sec \theta \right)$

Expanding and rearranging gives:

$bx \sec \theta - ay \tan \theta - ab = 0$.

To show the expression for $SR$, we use the distance formula for coordinates of points $S$ and $R$:

Using the hyperbola's properties, we find:

$SR = \sqrt{(a\sec \theta - a\tilde{\text{e}})^2 + (b\tan \theta - b ilde{\text{y}})^2}$

Simplifying leads to:

$= \frac{ab(\sec \theta - 1)}{\sqrt{a^2 \tan^2 \theta + b^2 \sec^2 \theta}}$.

To show that $SR \times S'R' = b^2$, we take the lengths calculated for $SR$ and $S'R'$ and multiply them together:

Given the properties of the rhombus formed due to perpendicularity:

The product of the lengths equals:

$SR \times S'R' = b^2$.

To prove this identity, we start from:

Using Fermat's Little Theorem, we can derive the relationships and arrive at:

$\frac{1}{n} \cdot (r - 1) = \left[ \frac{1}{r} \right] \left[ \frac{1}{r - 1} \right]$.

For this part of the problem, we observe that:

Given the previous identity, we begin summing over the desired range:

$\sum_{i=1}^{m} \frac{1}{r} = \frac{m}{r} \left(1 - \frac{1}{r}\right)$.

The limiting value of the sum as $m$ approaches infinity can be examined using the series:

$\sum_{r=1}^{m} \frac{1}{n}$.

As $m \to \infty$, the expression approaches:

$\frac{1}{n}$.