Ronald Hui Tak Sun Secondary School HKDSE Mathematics Ronald Hui Tak Sun Secondary School

Homework SHW6-C1 Sam L SHW6-R1 Walter (RD) Ronald HUI

Homework SHW7-01 SHW7-B1 SHW7-C1 SHW7-R1 SHW7-P1 Pako Sam L Daniel (RD), Pako SHW7-R1 Kelvin, Sam L, Pako SHW7-P1 Sam L, Pako, Walter Ronald HUI

Standard Quiz Result of Standard Quiz Major Problems Full mark: 30 Highest: 20 (2) Lowest: 0 (x3) – Same as last year! Average: 6.27 (2.05) Number of students (10-14): 2 Number of students (0 – 9): 17 Major Problems Forget how to use the equation Did not start to read the questions Just give up!

Standard Quiz Result of Standard Quiz 4th: CHAN Chun Hang, Jason (NEW) HUI Ka Ho, Ken (NEW) 3rd: WONG Vincent Wai Shun (1) 1st: CHEUNG Ka Chun, Nathan (NEW) WONG Yin Man Marco (=) Congratulations!!! Well done!!!

Standard Quiz (24) Result of Standard Quiz Major Problems Full mark: 24 Highest: 22 (1) Lowest: 3 (1) Average: 10.55 (1.67) Number of students (10-11): 1 Number of students (0 – 9): 11 Major Problems Cannot understand well the questions Did not start to try

Standard Quiz (24) Result of Standard Quiz (24) 10th: WONG Yin Man Marco (NEW) 7th: CHAN Chun Hang, Jason (=) WONG Chi Him Aston (NEW) YEUNG Ho Sing, Stitch (2) 5th: CHEONG Shun Yin, Matthew (4) YAN Tin Lok (4) 4th: WAN Hang Nam, Hanki (1) 3rd: WONG Vincent Wai Shun (NEW) 2nd: YEUNG Choi Sum, Sam (3) 1st: CHEUNG Ka Chun, Nathan (=) Congratulations!!!

Coming up SHW8-A1: Deadline past!!!!! Ronald HUI

Algebraic Equation of a Locus

Consider a point P which moves in such a way that it maintains a fixed distance of 2 units from a fixed point A. y By introducing a rectangular coordinate system, the locus of P can be described by an algebraic equation. This equation is called the equation of the locus. A P 2 units x (0, 0) Besides using words, we can use an equation to describe the locus of a moving point. The equation of the locus can be found by the following steps.

Step 1: Let the coordinates of a moving point be (x, y). y Let (x, y) be the coordinates of P. A P 2 units (x, y) x (0, 0)

Step 2: Set up an equation connecting x and y according to the condition(s) which the moving point must satisfy. y Let (x, y) be the coordinates of P. ∵ A P 2 units (x, y) ∴ By the distance formula x (0, 0)

Step 3: Simplify the equation if possible. y Let (x, y) be the coordinates of P. ∵ A P 2 units (x, y) ∴ ∴ The equation of the locus of P is . x (0, 0) From the equation, we can see that the required locus is a circle centred at the origin.

In general, the equation of the locus can be found by the following steps: Let the coordinates of a moving point be (x, y). Step 2: Set up an equation connecting x and y according to the condition(s) which the moving point must satisfy. Step 3: Simplify the equation if possible. Let us look at more examples. In general, we can know the location and the shape of the locus from the equation of the locus.

The figure shows two points A(3, –1) and B(–2, 1) The figure shows two points A(3, –1) and B(–2, 1). P is a moving point such that PA = PB. Find the equation of the locus of P. A(3, –1) B(–2, 1) P y x

Step 1: Let the coordinates of a moving point be (x, y). A(3, –1) B(–2, 1) P y x (x, y) Step 1 Let (x, y) be the coordinates of P.

Step 2: Set up an equation connecting x and y according to the condition(s) which the moving point must satisfy. A(3, –1) B(–2, 1) P y x (x, y) In this case, the condition that must be satisfied is PA = PB. Let (x, y) be the coordinates of P. Step 2 Use the distance formula to set up the equation connecting x and y.

Step 3: Simplify the equation if possible. A(3, –1) B(–2, 1) P y x (x, y) It is the perpendicular bisector of AB. Let (x, y) be the coordinates of P. Step 3 ∴ The equation of the locus of P is 10x – 4y – 5 = 0. From the equation, we can see that the locus is a straight line.

The figure shows the point E(0, –5) and the straight line L: y = 2. (a) Find the equation of the locus of a moving point P such that it is equidistant from E and L. (b) What kind of curve is the locus in (a)? y L: y = 2 x P E(0, –5)

Step 1: Let the coordinates of a moving point be (x, y). y L: y = 2 F x Step 1 (a) Let (x, y) be the coordinates of P, and F be a point on L such that PF ⊥ L. P(x, y) E(0, –5)

Step 2: Set up an equation connecting x and y according to the condition(s) which the moving point must satisfy. y L: y = 2 F x (a) Let (x, y) be the coordinates of P, and F be a point on L such that PF ⊥ L. P(x, y) E(0, –5) Step 2 PF is a vertical line segment.

Step 3: Simplify the equation if possible. y L: y = 2 F x (a) Let (x, y) be the coordinates of P, and F be a point on L such that PF ⊥ L. P(x, y) E(0, –5) This is an equation of a parabola. Step 3 (b) The locus of P is a parabola.

Follow-up question In the figure, A(–2, –4) and B(3, –2) are two points, and P is a moving point. (a) Find the equation of the locus of P such that AP ⊥ BP. (b) What kind of curve is the locus in (a)? y x B(3, –2) A(–2, –4) P

Step 2: Set up an equation connecting x and y according to the condition(s) which the moving point must satisfy. Step 1: Let the coordinates of a moving point be (x, y). Step 3: Simplify the equation if possible. A(–2, –4) B(3, –2) y x (a) Let (x, y) be the coordinates of P. ∵ AP ⊥ BP ∴ The product of the slopes of two perpendicular lines is –1. P(x, y) This is an equation of a circle. ∴ The equation of the locus of P is x2 + y2 – x + 6y + 2 = 0.

A(–2, –4) B(3, –2) y x (b) The locus of P is a circle with AB as diameter, excluding A and B. locus of P

Let us look at the following example. Sometimes, we can make use of the geometric properties to find the equation of locus.

A moving point P maintains an equal distance from two parallel lines L1: y = x + 5 and L2: y = x + 9. Find the equation of the locus of P. The required locus is the straight line parallel to and midway between L1 and L2. We name the line L3. y x O M A B R S L2 L3 L1 With the notations in the figure, ∵ y-intercept of L1 = 5 and y-intercept of L2 = 9 ∴ Coordinates of A = (0, 5) and coordinates of B = (0, 9)

∴ Coordinates of M y x O M A B R S L2 L3 L1 ∴ y-intercept of L3 = 7 ∵ △MRA △MSB (ASA) ∴ (corr. sides, △s) Slope of L3 = slope of L1 = 1 ∴ The equation of L3 is y = x + 7 ∴ The equation of the locus of P is y = x + 7.

Follow-up question Find the equation of the locus of a moving point P such that P maintains a fixed distance of 3 units from the line L: 3x + 1 = 0. locus of P y 3 units 3 units ∵ L is a vertical line with x-intercept . ∴ The required locus is a pair of vertical lines at a distance of 3 units from L. x O

x-intercept of the vertical line 3 units on the left of L locus of P y 3 units 3 units x-intercept of the vertical line 3 units on the right of L x O ∴ The required equations are and .