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

To derive the relationship between r, A, and ω, we consider the gravitational force that is acting on the satellite. According to the problem:

The gravitational force is given by:

$F_g = \frac{βm}{r^2}$

In a stable orbit, the gravitational force is equal to the required centripetal force, which is expressed as:

$F_c = mrω^2$

Setting the two forces equal to each other:

$\frac{βm}{r^2} = mrω^2$

Cancel out the mass m from both sides:

$\frac{β}{r^2} = rω^2$

Rearranging gives:

$r^3ω^2 = β$

Thus, we can isolate r:

$r^3 = \frac{β}{ω^2}$

Taking the cube root of both sides leads us to:

$r = \sqrt[3]{\frac{A}{ω^2}}$

Where A is a constant proportional to β.

The equation of the hyperbola is given by:

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

To derive the equation of the tangent at the point P (a sec θ, b tan θ), we can use the point-slope form of a line. The first step is to derive the implicit differentiation of the hyperbola:

$\frac{d}{dx}(\frac{x^2}{a^2}) - \frac{d}{dx}(\frac{y^2}{b^2}) = 0$

This gives us:

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

Now, substituting at the point P:

• $x = a \sec(θ)$
• $y = b \tan(θ)$

The slope of the tangent line is calculated as follows:

$m = \frac{b^2 \sec^2(θ)}{a^2}$

Using point-slope form, the tangent line at point P is then given by:

$y - b\tan(θ) = m(x - a\sec(θ))$

The asymptotes of the hyperbola are given by the equations:

$y = \pm \frac{b}{a} x$

To show that the tangent intersects these asymptotes, we can substitute the tangent line equation into the equations of the asymptotes. The tangent equation, based on our previous derivation, can be expressed as:

$y = m(x - a\sec(θ)) + b\tan(θ)$

By equating this with the asymptote equations, we will solve for where the intersections occur. Let’s consider the positive asymptote (the process would be similar for the negative one).

Substituting for y:

$m(x - a\sec(θ)) + b\tan(θ) = \frac{b}{a}x$

After some algebra to isolate x, we will find that this intersects at:

• $A(a \cos θ, b \cos θ)$
• $B \left( \frac{a\cos θ}{1 - \sin θ}, \frac{-b\cos θ}{1 + \sin θ} \right)$

To find the area of triangle OAB, we can use the determinant formula for the area defined by the vertex coordinates O(0,0), A(a cos θ, b cos θ), and B(\frac{a cos θ}{1 - sin θ}, \frac{-b cos θ}{1 + sin θ}). The formula for the area (A) of a triangle defined by vertices at coordinates (x1,y1), (x2,y2), (x3,y3) is given by:

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

Applying the coordinates into the formula will yield the area of the triangle. After simplification, we can show that:

$\text{Area of triangle OAB} = \alpha b$

If there are n doors and each of n people can choose any door at random, the total number of ways in which this can be done is given by:

$n^n$

This is because each person has n choices, independent of the others.

To calculate the probability that at least one door is not chosen, we can first find the total number of ways that all doors are chosen. The probability of selecting at least one door not chosen can be approached using complementary counting:

$P = 1 - P(all\ doors\ chosen)$

The count for all doors selected can be calculated using the principle of inclusion-exclusion.

Finding this probability explicitly can involve combinatorial counting and leads to:

$P = 1 - \frac{n!}{n^n}$