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.

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Asked on April 5, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-04-05T01:00:46.681Z

Here are the solutions to Question 5. Question 5 (i): The parametric equations of a curve are given by x = 1 + t and y = 2 + (1)/(t), where t ≠ 0. Part 1: Show that an equation of the normal to the curve at the point where t=2 is 8x - 2y - 19 = 0. Step 1: Find the coordinates of the point on the curve when t=2. Substitute t=2 into the parametric equations: x = 1 + 2 = 3 y = 2 + (1)/(2) = (5)/(2) The point is (3, (5)/(2)). Step 2: Find the gradient of the tangent, (dy)/(dx). First, find (dx)/(dt) and (dy)/(dt): (dx)/(dt) = (d)/(dt)(1+t) = 1 (dy)/(dt) = (d)/(dt)(2 + t^-1) = -t^-2 = -(1)/(t^2) Now, use the chain rule (dy)/(dx) = (dy/dt)/(dx/dt): (dy)/(dx) = (-1/t^2)/(1) = -(1)/(t^2) Step 3: Find the gradient of the tangent at t=2. Substitute t=2 into (dy)/(dx): m_tangent = -(1)/((2)^2) = -(1)/(4) Step 4: Find the gradient of the normal at t=2. The gradient of the normal, m_normal, is the negative reciprocal of the tangent's gradient: m_normal = -(1)/(m_tangent) = -(1)/((-1/4)) = 4 Step 5: Find the equation of the normal. Using the point-slope form y - y_1 = m(x - x_1) with the point (3, (5)/(2)) and m_normal = 4: y - (5)/(2) = 4(x - 3) y - (5)/(2) = 4x - 12 Multiply the entire equation by 2 to eliminate the fraction: 2y - 5 = 8x - 24 Rearrange the terms to match the given equation form: 0 = 8x - 2y - 24 + 5 8x - 2y - 19 = 0 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. Step 1: Substitute the parametric equations of the curve into the equation of the normal. The curve is x = 1 + t and y = 2 + (1)/(t). The normal equation is 8x - 2y - 19 = 0. Substitute x and y: 8(1 + t) - 2(2 + (1)/(t)) - 19 = 0 Step 2: Simplify and solve the equation for t. 8 + 8t - 4 - (2)/(t) - 19 = 0 8t - (2)/(t) + (8 - 4 - 19) = 0 8t - (2)/(t) - 15 = 0 Multiply the entire equation by t (since t ≠ 0): 8t^2 - 2 - 15t = 0 Rearrange into a standard quadratic equation form: 8t^2 - 15t - 2 = 0 Step 3: Solve the quadratic equation for t. Factor the quadratic equation: 8t^2 - 16t + t - 2 = 0 8t(t - 2) + 1(t - 2) = 0 (8t + 1)(t - 2) = 0 This gives two possible values for t: t - 2 = 0 t = 2 8t + 1 = 0 t = -(1)/(8) 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 -(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. Step 1: Determine the total number of staff and the gender distribution for equity. 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. Step 2: Consider the two possible cases for the Head Teacher's gender. 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 31 = 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: 43 = (4!)/(3!1!) = 4 ways. Number of ways to choose 4 females from 6 female applicants: 64 = (6!)/(4!2!) = (6 × 5)/(2 × 1) = 15 ways. Total ways for Case 1 = (Ways to choose male HT) × (Ways to choose 3 males) × (Ways to choose 4 females) = 3 × 4 × 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 21 = 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: 44 = 1 way. Number of ways to choose 3 females from 6 female applicants: 63 = (6!)/(3!3!) = (6 × 5 × 4)/(3 × 2 × 1) = 20 ways. Total ways for Case 2 = (Ways to choose female HT) × (Ways to choose 4 males) × (Ways to choose 3 females) = 2 × 1 × 20 = 40 Step 3: Calculate the total number of different ways. Total ways = Ways for Case 1 + Ways for Case 2 = 180 + 40 = 220 The number of different ways of selecting the staff of 8 if gender equity must be respected is 220.

Accepted Answer · 2026-04-05T01:00:46.681Z

Here are the solutions to Question 5. Question 5 (i): The parametric equations of a curve are given by x = 1 + t and y = 2 + (1)/(t), where t ≠ 0. Part 1: Show that an equation of the normal to the curve at the point where t=2 is 8x - 2y - 19 = 0. Step 1: Find the coordinates of the point on the curve when t=2. Substitute t=2 into the parametric equations: x = 1 + 2 = 3 y = 2 + (1)/(2) = (5)/(2) The point is (3, (5)/(2)). Step 2: Find the gradient of the tangent, (dy)/(dx). First, find (dx)/(dt) and (dy)/(dt): (dx)/(dt) = (d)/(dt)(1+t) = 1 (dy)/(dt) = (d)/(dt)(2 + t^-1) = -t^-2 = -(1)/(t^2) Now, use the chain rule (dy)/(dx) = (dy/dt)/(dx/dt): (dy)/(dx) = (-1/t^2)/(1) = -(1)/(t^2) Step 3: Find the gradient of the tangent at t=2. Substitute t=2 into (dy)/(dx): m_tangent = -(1)/((2)^2) = -(1)/(4) Step 4: Find the gradient of the normal at t=2. The gradient of the normal, m_normal, is the negative reciprocal of the tangent's gradient: m_normal = -(1)/(m_tangent) = -(1)/((-1/4)) = 4 Step 5: Find the equation of the normal. Using the point-slope form y - y_1 = m(x - x_1) with the point (3, (5)/(2)) and m_normal = 4: y - (5)/(2) = 4(x - 3) y - (5)/(2) = 4x - 12 Multiply the entire equation by 2 to eliminate the fraction: 2y - 5 = 8x - 24 Rearrange the terms to match the given equation form: 0 = 8x - 2y - 24 + 5 8x - 2y - 19 = 0 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. Step 1: Substitute the parametric equations of the curve into the equation of the normal. The curve is x = 1 + t and y = 2 + (1)/(t). The normal equation is 8x - 2y - 19 = 0. Substitute x and y: 8(1 + t) - 2(2 + (1)/(t)) - 19 = 0 Step 2: Simplify and solve the equation for t. 8 + 8t - 4 - (2)/(t) - 19 = 0 8t - (2)/(t) + (8 - 4 - 19) = 0 8t - (2)/(t) - 15 = 0 Multiply the entire equation by t (since t ≠ 0): 8t^2 - 2 - 15t = 0 Rearrange into a standard quadratic equation form: 8t^2 - 15t - 2 = 0 Step 3: Solve the quadratic equation for t. Factor the quadratic equation: 8t^2 - 16t + t - 2 = 0 8t(t - 2) + 1(t - 2) = 0 (8t + 1)(t - 2) = 0 This gives two possible values for t: t - 2 = 0 t = 2 8t + 1 = 0 t = -(1)/(8) 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 -(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. Step 1: Determine the total number of staff and the gender distribution for equity. 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. Step 2: Consider the two possible cases for the Head Teacher's gender. 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 31 = 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: 43 = (4!)/(3!1!) = 4 ways. Number of ways to choose 4 females from 6 female applicants: 64 = (6!)/(4!2!) = (6 × 5)/(2 × 1) = 15 ways. Total ways for Case 1 = (Ways to choose male HT) × (Ways to choose 3 males) × (Ways to choose 4 females) = 3 × 4 × 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 21 = 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: 44 = 1 way. Number of ways to choose 3 females from 6 female applicants: 63 = (6!)/(3!3!) = (6 × 5 × 4)/(3 × 2 × 1) = 20 ways. Total ways for Case 2 = (Ways to choose female HT) × (Ways to choose 4 males) × (Ways to choose 3 females) = 2 × 1 × 20 = 40 Step 3: Calculate the total number of different ways. Total ways = Ways for Case 1 + Ways for Case 2 = 180 + 40 = 220 The number of different ways of selecting the staff of 8 if gender equity must be respected is 220.

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.

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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.

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