A small bead of mass m is attached to one end of a light string of length R - HSC - SSCE Mathematics Extension 2 - Question 5 - 2011 - Paper 1

To show that $N = \frac{1}{2} m \, \sec{\theta} - \frac{1}{2} m \frac{R}{r} \, \csc{\theta},$ we start from the vertical resolution of forces.

1. From the equation, we solve for N:

   • Isolate N in terms of F and other known quantities from the equations above.
   • By substituting the expression for F using the equations, we relate the terms accordingly.

2. Simplifying further will lead to the required form.

To verify the condition:

1. Use the derived formula for N and set up the inequality for it to remain non-negative: $N \geq 0.$
2. Substitute the expression for N obtained previously and solve the inequality for ω:
   • Recognizing that for contact maintenance, we need to ensure the gravitational influence balances with the angular momentum derived forces
   • Hence deducing the required relationship: if ( \omega \leq \sqrt{\frac{g}{h}} ), the bead remains in contact.

We start with: $\frac{p}{1 + p} + \frac{q}{1 + q} + \frac{r}{1 + r}.$

1. Applying the common denominator, expand: $\Rightarrow \frac{p(1 + q)(1 + r) + q(1 + p)(1 + r) + r(1 + p)(1 + q)}{(1 + p)(1 + q)(1 + r)}.$
2. Simplifying will yield:
   • Look carefully at how all interactions occur in terms of the positivity condition and arrive at the conclusion.

Using the reflection property of ellipses, which states that any ray from one focus reflects off the tangent line to the ellipse and passes through the other focus:

1. At point P, any light ray coming from focus S reflects such that it passes through focus S'.
2. Thus, we have:
   • The segments ( SQ ) and ( RQ ) are equal, confirming the property.

From the ellipse definition, the distance between the foci is given by:

• Using the properties of ellipses, the distance S'S relates to the semi-major axis a: $S'R = 2a.$ This follows from properties of elliptical geometry that state distances involving foci and points on the ellipse return measures directly relative to semi-major axis.

To show that point Q lies on the circle defined by the equation:

1. Connect the properties established about distances from P, S, and S' to point Q.
2. Demonstrate that under the relationships established, Q maintains a radial distance from the origin equivalent to a, thus proving: $x^{2} + y^{2} = a^{2}.$