Chapter 8.1 Conic Sections/ParabolasHonors Pre-CalculusRogers High School

Introduction to Conic Sections

Conic sections are defined geometrically as the result of the intersection of a plane with a right circular cone.

Algebraically, conic sections are second degree equations of two variables which includes equations of the form 𝐴𝑥2 + 𝐵𝑥𝑦 + 𝐶𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 where A, B, and C do not all equal 0. The type of conic section is determined by the value of the coefficients.

The most commonly studied conic sections include parabolas, ellipses, circles, lines, and points. For the next 2 or 3 weeks, we will focus on parabolas, circles, ellipses, and hyperbolas.

Graphical Representation of Conic Sections

Parabolas (Basics)

Parabolas are defined as follows:

Parabola Terminology

Focus – a fixed point which lies on the axis of symmetry and “inside” the parabola

Directrix – a fixed line which is perpendicular to the axis of symmetry and lies “outside” the parabola

Vertex – the point where the parabola and the axis of symmetry intersect; it is halfway between the focus and directrix

Focal Axis or Axis of Symmetry – a line which passes through the middle of a parabola and intersects the focus, vertex, and directrix

Parabola Terminology

Focal Length – the directed distance from the vertex to the focus of a parabola

Focal Width – the length of a chord which is perpendicular to the focal axis and passes through the focus of a parabola

Latus Rectum – a chord which is perpendicular to the axis and passes through the focus of the parabola

Standard Form (Equation) – equations of the form 𝑥 − ℎ 2 = 4𝑝(𝑦 − 𝑘) or 𝑦 − 𝑘 2 = 4𝑝(𝑥 − ℎ)

Parabolas with Vertex at (0,0)

Graphs of Parabolas with Vertex at Origin

EXAMPLE #1

Find the focus, directrix, and the focal width of the parabola defined by the

equation: 𝑦 =−1

3𝑥2

First, we need this to be in standard form. So we have 𝑥2 = −3𝑦.

We need the value of p. From the equation, we know 4𝑝 = −3. Thus, p is −3

4.

The focus is always at (0, p). So it is at (0, -3/4).

The directrix is located at 𝑦 = −𝑝. Thus, the directrix is at 𝑦 =3

4.

Finally, the focal width is 4𝑝 . Thus, the focal width is 3.

EXAMPLE #2

Find the equation, in standard form, of a parabola if the focus is (-2, 0) and the directrix is the line x = 2.

It may help to briefly sketch the elements given to visualize the graph.

From your sketch, you should see that the parabola opens to the left. Additionally, its vertex must be halfway between the focus and directrix. So the vertex is at (0, 0).

We need to find p. We know p is the directed distance from the vertex to the focus. So p is -2.

Thus, our equation is of the form 𝑦2 = 4𝑝𝑥 so y2 = −8𝑥 is the equation.

HOMEWORK ASSIGNMENT (DAY 1)

p. 587 [7 – 10, 11 – 16, 17 – 20, 31, 32, 37, 39]

Parabolas with Vertex (h, k)

Graphs of Parabolas with Vertex (h, k)

EXAMPLE #3

Find the standard form equation of a parabola with vertex (3, 4) and focus (5, 4).

By graphing the information given, we can see this parabola opens to right. So the equation is of the form 𝑦 − 𝑘 2 = 4𝑝(𝑥 − ℎ) and p is positive as well.

We know the vertex is (3, 4). Thus, h = 3 and k = 4.

Finally, we need to find p. Recall that p is the directed distance from vertex to focus. So we are moving from x = 3 to x = 5. That is a distance of 2. So p = 2.

𝑦 − 𝑘 2 = 4𝑝(𝑥 − ℎ) becomes 𝑦 − 4 2 = 8(𝑥 − 3)

EXAMPLE #4

Use your calculator to graph 𝑦 − 4 2 = 8(𝑥 − 3) which was the equation we just found.

Notice that this equation will contain 𝑦2 which cannot normally be graphed in your standard Y = screen. So let’s work on solving for y.

𝑦 − 4 2 = 8 𝑥 − 3 [𝑂𝑅𝐼𝐺𝐼𝑁𝐴𝐿]

𝑦 − 4 = ± 8 𝑥 − 3 [𝑆𝑄𝑈𝐴𝑅𝐸 𝑅𝑂𝑂𝑇 𝐵𝑂𝑇𝐻 𝑆𝐼𝐷𝐸𝑆]

𝑦 = 4 ± 8 𝑥 − 3 [𝐴𝐷𝐷 4 𝐵𝑂𝑇𝐻 𝑆𝐼𝐷𝐸𝑆]

We can graph each part + and – separately in the calculator.

EXAMPLE #4

Put each part into the Y = screen. Then graph the result. You may need to change the window to get it to work well. Here I am using x values from -1 to 7 and y values from -2 to 10.

EXAMPLE #5

Show that 𝑦2 − 6𝑥 + 2𝑦 + 13 = 0 is a parabola and find its vertex, focus, and directrix.

First of all, this is a parabola because it has only one variable squared. However, let’s put it into standard form to see that as well.

𝑦2 − 6𝑥 + 2𝑦 + 13 = 0

𝑦2 + 2𝑦 + 1 = 6𝑥 − 13 + 1 [𝐶𝑂𝑀𝑃𝐿𝐸𝑇𝐸 𝑆𝑄𝑈𝐴𝑅𝐸]

𝑦 + 1 2 = 6𝑥 − 12 𝐹𝐴𝐶𝑇𝑂𝑅

𝑦 + 1 2 = 6 𝑥 − 2 [𝐹𝐴𝐶𝑇𝑂𝑅]

EXAMPLE #5

𝑦 + 1 2 = 6 𝑥 − 2

Standard form is 𝑦 − 𝑘 2 = 4𝑝 𝑥 − ℎ

Thus, we know the vertex is (h, k) which is (2, -1).

The value of p must be 6/4 or 3/2.

The focus would be at (h + p, k) which is (3.5, -1).

The directrix is the line x = h – p which would be x = 0.5.

Applications of Parabolas

Since everyone wonders “Why do we need to know this?” I will point out the following reasons that parabolas are useful. Some of these may be known already.

Projectile Motion – Generally, projectiles follow parabolic curves when it motion

Physics – Many physics applications such as the stopping distance of a car use quadratic functions

Electromagnetic Waves – Many products which rely on electromagnetic waves such as car headlights, some headers, high tech microphones, and satellites use parabolic shapes to maximize effectiveness

EXAMPLE #6

On the sidelines of each of its televised football games, the ESPN uses a parabolic reflector with a microphone at the reflector’s focus to capture the conversations among players on the field. If the parabolic reflector is 3 ft across and 1 ft deep, where should the microphone be placed?

First of all, we will sketch a 2 dimension representation of the parabolic curve being used here.

EXAMPLE #6

From our drawing, we have a parabola of the form 𝑥2 = 4𝑝𝑦.

We know the points (-1.5, 1) and (1.5, 1) lie on the graph as we were told it was 3 feet across.

We can use these points to find p.

±1.5 2 = 4𝑝 1 𝑆𝑈𝐵𝑆𝑇𝐼𝑇𝑈𝑇𝐼𝑂𝑁

2.25 = 4𝑝 𝑆𝐼𝑀𝑃𝐿𝐼𝐹𝑌

0.5625 = 𝑝 [𝐷𝐼𝑉𝐼𝐷𝐸 𝐵𝑌 4]

So the microphone should be 0.5625 feet from the vertex of the reflector.

HOMEWORK ASSIGNMENT (DAY 2)

p. 587 [1 – 6, 21 – 29 odd, 33 – 36, 49 – 55 odd, 61]