A particle P of mass m moves with constant angular velocity ω on a circle of radius r - HSC - SSCE Mathematics Extension 2 - Question 4 - 2003 - Paper 1

The gravitational force acting on the satellite is given by:

$F_g = \frac{Am}{r^2}$

This force must equal the centripetal force required to keep the satellite in circular motion:

$F_c = m \omega^2 r$

Setting these forces equal provides:

$\frac{Am}{r^2} = m \omega^2 r$

By canceling $m$ from both sides and rearranging, we have:

$A = \omega^2 r^3$

From here, solving for r yields:

$r^3 = \frac{A}{\omega^2}$

Thus,

$r = \sqrt[3]{\frac{A}{\omega^2}}$

To derive the equation of the tangent to the hyperbola given by:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

at point P $(a \sec \theta, b \tan \theta)$, we utilize the point-slope form of the tangent line to a hyperbola. The slope ($m$) of the tangent at P can be found using implicit differentiation:

1. Differentiate both sides:

   • The derivative of left-hand side with respect to x gives:

   $\frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0$

   • Solving for $\frac{dy}{dx}$ gives:

   $\frac{dy}{dx} = \frac{y \cdot a^2}{x \cdot b^2}$

2. Substitute point P into slope formula to find the slope at that point.

Once the slope is determined, use the point-slope form:

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

To show where the tangent intersects the asymptotes of the hyperbola:

1. The asymptotes of the hyperbola are:

   • $y = \pm \frac{b}{a} x$

2. Substitute the equation of the tangent into the equation of the asymptotes to find the points of intersection.

Using the derived tangent equation from part (i), set it equal to the asymptotes' equations to isolate x and y coordinates leading to the points A and B as outlined.

To find the area of triangle OAB, we can use the formula for the area of a triangle based on coordinates:

$\text{Area} = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |$

1. Let O be the origin (0,0), and A and B be the coordinates as derived previously.

2. Substitute these coordinates into the area formula.

3. After performing the calculations and simplifying, it can be shown:

$\text{Area} = \alpha b$

When n people choose a door to enter the hall, and each person can choose independently, the total number of ways this can be done is calculated as follows:

Each person has n choices (doors). Thus, the total combinations can be expressed as:

$n^n$

To find the probability that at least one door will not be chosen by any of the n people, it is often easier to use the complement rule:

1. Calculate the probability that all doors are chosen by at least one person, denoted as $P(all)$.

2. The probability that at least one door is not chosen is then:

$P(at least \, one \, not \, chosen) = 1 - P(all)$

3. $P(all)$ can be derived using counting principles and depends on combinations calculated previously.