Conic Sections – SS3 Further Mathematics Lesson Note (Second Term)

A parabola is the locus of points that are equidistant from a fixed point, called the focus, and a fixed line, called the directrix. A parabola with its vertex at the origin has the equation y² = 4ax, with focus (a, 0) and directrix x = -a.

Behavioural objectives

By the end of the lesson, students should be able to recognise the two-dimensional equations of:

1. a parabola;
2. an ellipse;
3. a hyperbola; and

and solve problems on a parabola, an ellipse and a hyperbola.

Instructional resources

• Solid shapes of parabolic, elliptic and hyperbolic types, and similar items.

Lesson presentation

The teacher presents the lesson step by step. The teacher first asks students questions based on previous lessons, for example, “What is a circle?”

Step 1: Identification of prior knowledge

Mode: Entire class.

The teacher arouses the students’ interest by asking questions relevant to the lesson, for example, “What is a circle?”

What is a parabola?

A parabola is the locus of points equidistant from a given point, called the focus, and from a given line, called the directrix.

Parts of a parabola

From the diagram above:

• Line AB, a distance of a from the y-axis, is called the directrix.
• Line AF is called the axis of symmetry.
• Point V is called the vertex.
• The line with endpoints R and Q is called the latus rectum.

Equation of the parabola

The equation of the parabola is y² = 4ax.

If the vertex of the parabola y² = 4ax is translated to the point (x₁, y₁), the equation of the corresponding parabola becomes:

(y – y₁)² = 4a(x – x₁)

This equation is said to be in the standard or canonical form.

Worked examples

Example 1

Find the focus and directrix of the parabola y² = 16x.

Solution

Compare y² = 16x with y² = 4ax.

4a = 16, so a = 4.

Hence, the focus is (4, 0) and the directrix is x = -4.

Example 2

Write the equation y² – 4y – 12x + 40 = 0 in canonical form and hence determine the vertex, focus and directrix.

Solution

1. y² – 4y – 12x + 40 = 0
2. y² – 4y + 4 – 12x + 36 = 0
3. y² – 4y + 4 = 12x – 36
4. (y – 2)² = 12(x – 3)

Hence, the vertex is (3, 2).

Since 4a = 12, a = 3.

The focus is (3 + 3, 2) = (6, 2).

The directrix is x = 3 – 3 = 0.

Example 3

Find the equation of the parabola whose vertex is the origin and whose focus is the point F(5, 0).

Solution

Let the equation of the parabola be y² = 4ax.

The focus F(a, 0) = (5, 0), so a = 5.

Hence, the equation of the parabola is y² = 20x.

Evaluation

1. Find the foci and directrices of the following parabolas:
   • (a) y² = 32x. Answer: focus (8, 0), directrix x = -8.
   • (b) y² = -40x. Answer: focus (-10, 0), directrix x = 10.
   • (c) x² = 12y. Answer: focus (0, 3), directrix y = -3.
   • (d) x² = -50y. Answer: focus (0, -25/2), directrix y = 25/2.
2. Write the equations of the following parabolas in canonical form and hence determine their vertices, foci and directrices:
   • (a) y² – 6y – 2x + 19 = 0. Answer: (y – 3)² = 2(x – 5), V(5, 3), F(11/2, 3) and x = 9/2.
   • (b) y² + 4y + 4x + 16 = 0. Answer: (y + 2)² = -4(x + 3), V(-3, -2), F(-4, -2) and x = -2.
   • (c) x² – 6x – 4y + 13 = 0. Answer: (x – 3)² = 4(y – 1), V(3, 1), F(3, 2) and y = 0.
   • (d) x² + 4x + 4y + 16 = 0. Answer: (x + 2)² = -4(y + 3), V(-2, -3), F(-2, -4) and y = -2.

Assignment

Find the equations of the following parabolas, whose vertex is the origin and whose foci are the points:

1. F(-4, 0)
2. F(0, 5)
3. F(0, -4)