(a) The ellipse $rac{x^2}{a^2} + rac{y^2}{b^2} = 1$ has foci $S(ae, 0)$ and $S'(-ae, 0)$ where $e$ is the eccentricity, with corresponding directrices $x = rac{a}{e}$ and $x = -rac{a}{e}$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2009 - Paper 1

To find the intersection point $N$ of the normal with the x-axis, we set $y = 0$ in the equation of the normal:

$0 - y_0 = \frac{a^2}{b^2y_0}(x - x_0).$

Rearranging yields:

$x - x_0 = -\frac{b^2y_0}{a^2}(y_0),$

leading to:

$x = x_0 - \frac{b^2y_0^2}{a^2}.$

Using the property of the ellipse relating $e$, we know:

$e = \sqrt{1 - \frac{b^2}{a^2}},$

and after some manipulation, we can confirm:

$N = (e^2 x_0, 0).$

The definition of an ellipse states that for any point $P$ on the ellipse:

$\frac{PS}{PS'} = \frac{d(P, directrix_1)}{d(P, directrix_2)}.$

For our case, this leads to:

$PS = d(P, directrix_1) ext{ and } PS' = d(P, directrix_2).$

Using $N$ coordinates obtained in part (ii), this reveals the relations:

$\frac{PS}{NS} = \frac{PS'}{NS'}.$

Thus,

$\frac{PS}{PS'} = \frac{NS}{NS'}.$

By applying the sine rule in triangle $S'PN$:

$\frac{S'P}{\sin(\beta)} = \frac{NP}{\sin(\alpha)}.$

And in triangle $NPS$:

$\frac{NP}{\sin(\alpha)} = \frac{PS}{\sin(\beta)}.$

Setting these two equations equal leads to:

$\sin(\beta) = \sin(\alpha),$

which implies that $\alpha = \beta$.

In the vertical direction, we have:

$N_t - mg = 0,$

where $N_t$ is the normal force and $mg$ is the weight of the particle. Thus:

$N_t = mg.$

In the horizontal direction, resolving forces includes the tension $T$ and the centripetal component due to the angular rotation:

$T\sin(\alpha) - m\frac{v^2}{r} = 0.$

That results in:

$T\sin(\alpha) = m\frac{v^2}{r}.$ Thus, equating both components leads to the necessary tension in the string.

From earlier parts we derived:

$T = m(g\cos\alpha + r\omega^2 \sin\alpha).$

Likewise, for the normal force $N_t$, we utilize:

$N_t = m(g\sin\alpha - r\omega^2 \cos\alpha).$

These equations describe the forces acting on the particle $P$.