The parametric equations of a curve are given by x = 1 + t and y = 2 + (1)/(t), where t != 0. Show that an equation of the normal to the curve at the point where t=2 is 8x − 2y − 19 = 0.

Substitute $t=2$ into the parametric equations:

The point is $\left(3, \frac{5}{2}\right)$.

First, find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

Now, use the chain rule $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:

Substitute $t=2$ into $\frac{dy}{dx}$:

The gradient of the normal, $m_{\text{normal}}$, is the negative reciprocal of the tangent's gradient:

Using the point-slope form $y - y_1 = m(x - x_1)$ with the point $\left(3, \frac{5}{2}\right)$ and $m_{\text{normal}} = 4$:

Multiply the entire equation by 2 to eliminate the fraction:

Rearrange the terms to match the given equation form:

This matches the given equation, so it is shown.

Part 2: Find the value of $t$ at the point where this normal meets the curve again.

The curve is $x = 1 + t$ and $y = 2 + \frac{1}{t}$.
The normal equation is $8x - 2y - 19 = 0$.
Substitute $x$ and $y$:

Multiply the entire equation by $t$ (since $t \neq 0$):

Rearrange into a standard quadratic equation form:

Factor the quadratic equation:

This gives two possible values for $t$:

The value $t=2$ corresponds to the initial point where the normal was found. The question asks for the point where the normal meets the curve again.
Therefore, the other value of $t$ is the answer.

The value of $t$ at the point where this normal meets the curve again is $\boxed{-\frac{1}{8}}$.

Question 5 (ii):
A new primary school intends to recruit a head teacher and 7 other teachers.
There are 3 male and 2 female applicants for the position of head teacher.
There are 4 male and 6 female other applications for the teaching positions.
Find the number of different ways of selecting the staff of 8 if gender equity must be respected.

Total staff to be selected = 1 Head Teacher + 7 Other Teachers = 8 staff.
For gender equity with 8 staff, there must be 4 males and 4 females.

Case 1: The Head Teacher is Male.

• Number of ways to choose a male Head Teacher: There are 3 male applicants for Head Teacher, so $\binom{3}{1} = 3$ ways.
• Remaining staff to select: We need 7 other teachers. Since 1 male (Head Teacher) has been selected, we need 3 more males (4 total males - 1 male HT) and 4 females (4 total females) from the "Other Teachers" pool.
• Applicants for Other Teachers: 4 male, 6 female.
• Number of ways to choose 3 males from 4 male applicants: $\binom{4}{3} = \frac{4!}{3!1!} = 4$ ways.
• Number of ways to choose 4 females from 6 female applicants: $\binom{6}{4} = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15$ ways.
• Total ways for Case 1 = (Ways to choose male HT) $\times$ (Ways to choose 3 males) $\times$ (Ways to choose 4 females) $$= 3 \times 4 \times 15 = 180$$

Case 2: The Head Teacher is Female.

• Number of ways to choose a female Head Teacher: There are 2 female applicants for Head Teacher, so $\binom{2}{1} = 2$ ways.
• Remaining staff to select: We need 7 other teachers. Since 1 female (Head Teacher) has been selected, we need 4 males (4 total males) and 3 more females (4 total females - 1 female HT) from the "Other Teachers" pool.
• Applicants for Other Teachers: 4 male, 6 female.
• Number of ways to choose 4 males from 4 male applicants: $\binom{4}{4} = 1$ way.
• Number of ways to choose 3 females from 6 female applicants: $\binom{6}{3} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ ways.
• Total ways for Case 2 = (Ways to choose female HT) $\times$ (Ways to choose 4 males) $\times$ (Ways to choose 3 females) $$= 2 \times 1 \times 20 = 40$$

Total ways = Ways for Case 1 + Ways for Case 2

The number of different ways of selecting the staff of 8 if gender equity must be respected is $\boxed{220}$.