Conics D.Wetzel 2009. Parabolas Parabola: the set of points in a plane that are the same distance from a given point called the focus and a given line.

Conics

D.Wetzel

2009

Parabolas

Parabola: the set of points in a plane that are the same distance from a given point called the focus and a given line called the directrix.

Directrix

The light source is theFocus

The cross section of a headlightis an example of a parabola...

Here are some other examples of the parabola...

Directrix

Focus

d1

d1

d2

d2

d3

d3

Also, notice that the distance from the focus to any point on the parabola is equal to the distance from that point to the directrix...

We can determine the coordinates of the focus, and the equation of the directrix, given the equation of the parabola....

Vertex

Notice that the vertex is located at the midpoint between the focusand the directrix...

Standard Equation of a Parabola: (Vertex at the origin)

Equation Focus Directrix

ax2 = y (0, 1/4a) y = -1/4a

Equation Focus Directrix

ay2 = x (1/4a, 0) x = –1/4a

(If the x term is squared, the parabola goes up or down)

(If the y term is squared, the parabola goes left or right)

Example: Determine the focus and directrix of the parabola y = 4x2 :Since x is squared, the parabola goes up or down…

and the equation is: ax2 = y

To find the focus,

Let P represent the distance from

the vertex to the focus p = 1/4a

p = 1/(4)(4); p = 1/16

Focus: (0, p) Directrix: y = –p

Focus: (0, 1/16) Directrix: y = –1/16

See what this parabola looks like...

Example: Determine the focus and directrix of the parabola –3y2 – 12x = 0 :Since y is squared, the parabola goes left or right…

The equation is: ay2 = x

Solve for x: -3y2 = 12x

-3/12y2 = 12/12x

-1/4y2 = x

p = 1/4a so, p = 1/4(-1/4) = -1

Focus: (p, 0) Directrix: x = –p

Focus: (–1, 0)Directrix: x = 1

See what this parabola looks like...

Examples: Write the standard form of the equation of the parabola with focus at (0, 3) and vertex at the origin.

Since the focus is on the y axis, (and vertex at the origin)

the parabola goes up or down…

The equation is: ax2 = y

Since p = 3, 1/4a = 3 and a = 1/12

The standard form of the equation is: 1/12x2 = y

Example: Write the standard form of the equation of the parabola with directrix x = –1 and vertex at the origin.

Since the directrix is parallel to the y axis,(and vertex at the origin) the parabola goes left or right…The standard equation is: ay2 = x

Since p = 1, 1/4a = 1 and a = ¼

The standard form of the equation is: 1/4y2 = x

Circles

D.Wetzel

2009

circle: in a plane, the set of points equidistant from a given point, called the center.

radius: any segment whose endpoints are the center and a point on the circle.

If the circle is centered at (0, 0), and the radius is r, then the distance to any point, (x, y) on the circle (using the distance formula) is

x−0( )2+ y−0( )

2=r

x−0( )2+ y−0( )

2=r2

x2 +y2 =r2

square both sides...

now simplify...

This is the standard form of a circle with center (0,0) andradius r.

(x,y)

Example: Write an equation of the circle with its center at the origin with the point (–3, 4) on the circle.

Use the Standard Equation of the Circle to find the radius of the circle...

x2 +y2 =r2

−3( )2+ 4( )

2=r2

9+16=r2

25=r2

x2 +y2 =25

Ellipses

D.Wetzel

2009

Ellipses

Ellipse: set of all points in a plane such that the sum of the distances from two given points in a plane, called the foci, to any point on the curve is the same.

fociSum of the distances:

12 units

vertex vertex

co-vertex

co-vertex

The major axis is the line segment joining the vertices(through the foci)

The minor axis is the line segment joining the co-vertices (perpendicular to the major axis)

Standard Equation of an Ellipse (Center at Origin)

This is the equationif the major axis is horizontal.

The foci of the ellipse lie on the major axis, c units from the center, where c2 = a2 – b2

(c, 0) (–c, 0) (–a, 0) (a, 0)

(0, b)

(0, –b)

€

x2

a2 +y2

b2 =1; a > b

Standard Equation of an Ellipse (Center at Origin)

This is the equationif the major axis is vertical. (notice it looks the same)

The foci of the ellipse lie on the major axis, c units from the center, where c2 = a2 – b2

(0, c)

(0, –c)

(0, –b)

(0, b)

(a, 0) (–a, 0)

€

x2

a2 +y2

b2 =1; b> a

Example: Write an equation of the ellipse whose vertices are (0, –3) and (0, 3) and whose co-vertices are (–2, 0) and (2, 0). Find the foci of the ellipse. Since the major axis is

vertical, the equation isthe following:

Use c2 = b2 – a2 to find c. c2 = 32 – 22

c2 = 9 – 4 = 5c =

(0, c)

(0, –c)

(0, –3)

(0, 3)

(a, 0) (–2, 0)

5

Since b = 3 a = 2The equation is

The foci are

0, 5( ) and 0,− 5( )

€

x2

a2 +y2

b2 =1; b> a

€

x2

22 +y2

32=1

Example: Write the equation in standard form of 9x2 + 16y2 = 144. Find the foci and vertices of the ellipse.

Get the equation in standard form by dividing by 144:

Use c2 = a2 – b2 to find c. c2 = 42 – 32

c2 = 16 – 9 = 7c =

(c, 0) (–c,0)

(–4,0) (4, 0)

(0, 3)

(0,-3)

€

7

That means a = 4 b = 3

Vertices:Foci:Center: (0,0)

−4,0( ) and 4,0( )

Simplify...

− 7,0( ) and 7,0( )

€

9x2

144+16y2

144=144144

€

x2

16+

y2

9=1

Hyperbolas

D.Wetzel

2009

Hyperbolas

Hyperbola: set of all points such that the difference of the distances from any point to the foci is constant.

foci

vertices

asymptotes

Standard Equation of a Hyperbola (Center at Origin)

This is the equationif the transverse axis is horizontal. (notice it looks a lot like the equation of an ellipse.)

The foci of the hyperbola lie on the major axis, c units from the center, where c2 = a2+ b2

(c, 0) (–c, 0)

(–a, 0) (a, 0)

(0, b)

(0, –b)

€

x2

a2 −y2

b2 =1

Standard Equation of a Hyperbola (Center at Origin)

This is the equationif the transverse axis is vertical. (Notice that y is first for verticaland x was first for horizontal.)

The foci of the hyperbola lie on the major axis, c units from the center, where c2 = a2+ b2

(0, c)

(0, –c)

(0, –b)

(0, b) (a, 0) (–a, 0)

€

y2

b2 −x2

a2 =1

Example: Write the equation in standard form of 4x2 – 16y2 = 64. Find the foci and vertices of the hyperbola.

Divide both side of the equation by 64 to get the standard form:

Use c2 = a2 + b2 to find c. c2 = 42 + 22

c2 = 16 + 4 = 20c =

(c, 0) (–c,0)

(–4,0) (4, 0)

(0, 2)

(0,-2)

20=2 5

That means a = 4 b = 2

Vertices:Foci:

−4,0( ) and 4,0( )

Simplify...

−2 5,0( ) and 2 5,0( )

€

4x2

64−16y2

64=6464

€

x2

16−

y2

4=1

Example: Write an equation of the hyperbola whose foci are (0, –6) and (0, 6) and whose vertices are (0, –4) and (0, 4). Its center is (0, 0). Since the major axis is

vertical, the equation isthe following:

Since b = 4 and c = 6 , find a... c2 = b2+ a2

62 = 42 + a2

36 = 16 + a2

20 = a2

The equation of the hyperbola:

(–a, 0) (a, 0) (0, 4)

(0, –4)

(0, 6)

(0, –6)

€

y2

b2 −x2

a2 =1

€

y2

16−

x2

20=1

How do you graph a hyperbola?To graph a hyperbola, you need to know the center, the vertices, the co-vertices, and the asymptotes...

Draw a rectangle using +a and +b as the sides...

(5, 0) (–5,0) (–4,0) (4, 0) (0, 3)

(0,-3)

a = 4 b = 3

The asymptotes intersect at the center of the hyperbola and pass through the corners of a rectangle with corners (± a, ±b)

Example: Graph the hyperbola

c = 5

Draw the asymptotes (The asymptotes are just the diagonals through the box generated by a and b.)...

Draw the hyperbola...

€

x2

16−

y2

9=1

GRAPHS OF RATIONAL FUNCTIONSSTANDARD FORM OF EQUATIONS OF TRANSLATED CONICS

In the following equations the point (h, k) is the vertex of the parabola and the center of the other conics.

CIRCLE (x – h) 2 + (y – k)

2 = r 2

Horizontal Vertical

PARABOLA a(y – k)

2 + h = x a(x – h)

2 + k = y

HYPERBOLA (x – h)

2 (y – k)

2 – = 1b

2a 2

(y – k)

2 (x – h)

2 – = 1a

2b 2

ELLIPSE(x – h)

2 (y – k)

2 + = 1a

2 b 2 a > b

(x – h)

2 (y – k)

2 + = 1b

2 b > aa 2

WRITING AND GRAPHING EQUATIONS OF CONICS

Writing an Equation of a Translated Parabola

Write an equation of the parabola whose vertex is at (–2, 1) and whose focus is at (–3, 1).

SOLUTION

(–2, 1)(–2, 1)

Find h and k: The vertex is at (–2, 1),so h = – 2 and k = 1.

Plot what you have so far:

Because the parabola opens to the left,it has the form: x = a(y – k)2 + h

Giving you: x = a(y – 1)2 + -2

Writing an Equation of a Translated Parabola

Write an equation of the parabola whose vertex is at (–2, 1) and whose focus is at (–3, 1).

SOLUTION

(–3, 1)(–3, 1)(–2, 1)(–2, 1)

The standard form of the equation is x = -4(y – 1) 2 – 2 .

Find p: The distance between the vertex (–2, 1), and the focus (–3, 1) is

p = (–3 – (–2)) 2 + (1 – 1)

2 = 1

so p = 1 or p = – 1. Since p < 0, p = – 1.

p = 1/4a so a = -4

Graphing the Equation of a Translated Circle

Graph (x – 3) 2 + (y + 2)

2 = 16.

SOLUTION

Compare the given equation to the standard form of the equation of a circle:

(x – h) 2 + (y – k)

2 = r 2

You can see that the graph will be a circle with center at (h, k) = (3, – 2).

(3, – 2)

The radius is r = 4

Graphing the Equation of a Translated Circle

(3 + 4, – 2 + 0) = (7, – 2)

(3 + 0, – 2 + 4) = (3, 2) (3 – 4, – 2 + 0) = (– 1, – 2)

(3 + 0, – 2 – 4) = (3, – 6)

Draw a circle through the points.

Graph (x – 3) 2 + (y + 2)

2 = 16.

SOLUTION

(– 1, – 2)

(3, – 6)

(3, 2)

(3, – 2)

r

Plot several points that are each 4 units from the center:

(7, – 2)

Writing an Equation of a Translated Ellipse

Write an equation of the ellipse with foci at (3, 5) and (3, –1) and vertices at (3, 6) and (3, –2).

SOLUTION

Plot the given points and make a rough sketch.

(x – h) 2 (y – k)

2 + = 1b

2a 2

The ellipse has a vertical major axis,so its equation is of the form:

(3, 5)

(3, –1)

(3, 6)

(3, –2)Find the center: The center is halfwaybetween the vertices.

(3 + 3 6 + ( –2)2

(h, k) = , = (3, 2)2

SOLUTION

(3, 5)

(3, –1)

(3, 6)

(3, –2)

Writing an Equation of a Translated Ellipse

Find b: The value of b is the distancebetween the vertex and the center.

Find c: The value of c is the distancebetween the focus and the center.

b = (3 – 3) 2 + (6 – 2)

2 = 0 + 4 2 = 4

c = (3 – 3) 2 + (5 – 2)

2 = 0 + 3 2 = 3

Write an equation of the ellipse with foci at (3, 5) and (3, –1) and vertices at (3, 6) and (3, –2).

SOLUTION

(3, 5)

(3, –1)

(3, 6)

(3, –2)

Writing an Equation of a Translated Ellipse

Find a: Substitute the values of a and cinto the equation a

2 = b 2 – c

2 .

a 2 = 4

2 – 3 2

a 2 = 7

a = 7

167+ = 1The standard form is(x – 3) 2 (y – 2)

2

Write an equation of the ellipse with foci at (3, 5) and (3, –1) and vertices at (3, 6) and (3, –2).

Graphing the Equation of a Translated Hyperbola

Graph (y + 1) 2

– = 1.(x + 1)

2

4

SOLUTION

The y 2-term is positive, so the

transverse axis is vertical. Sincea

2 = 4 and b 2 = 1, you know that

a = 2 and b = 1.

Plot the center at (h, k) = (–1, –1). Plot the vertices 1 unit above and below the center at (–1, 0) and (–1, –2).

Draw a rectangle that is centered at (–1, –1) and is 2a = 2 units high and 2b = 4 units wide.

(–1, –2)

(–1, 0)

(–1, –1)

Graphing the Equation of a Translated Hyperbola

SOLUTION

Draw the asymptotes through the corners of the rectangle.

Draw the hyperbola so that it passes through the vertices and approachesthe asymptotes.

(–1, –2)

(–1, 0)

(–1, –1)

Graph (y + 1) 2

– = 1.(x + 1)

2

4

The y 2-term is positive, so the

transverse axis is vertical. Sincea

2 = 4 and b 2 = 1, you know that

a = 2 and b = 1.

Here are all of the conic section equations in standard form:

parabola: y = a(x – h)2 + k x = a(y – k)2 + h

circle: (x – h)2 + (y – k)2 = r2

ellipse: (x – h)2 + (y – k)2 = 1 a2 b2

hyperbola: (x – h)2 – (y – k)2 = 1 a2 b2

(y – k)2 – (x – h)2 = 1 b2 a2

How do you write an equation in standard form?Example: Write the equation of the ellipse in standard form

2x2 + 3y2 + 4x + 12y – 10 = 0

2(x + 1)2 + 3(y + 2)2 = 24

Group the x’s and y’s together...

2x2 + 4x +3y2 + 12y = 10 Factor out the GCF’s...2(x2 + 2x ) + 3(y2 + 4y ) = 10 Complete the square for each

variable.What will make each a perfect square trinomial?

+1 +4

Add the “real” amount to the other side (remember that they are being distributed)

+2 +12

Rewrite as the squares of binomials...Divide to set the right side equal to 1...

2(x + 1)2 + 3(y + 2)2 = 24 24 24 24

(x +1)2 + (y +2)2 = 1 12 8

Let’s try another…with graphing.

Example: Graph the hyperbola 9x2 – 4y2 + 18x + 16y – 43 = 0

9(x + 1)2 – 4(y – 2)2 =36

Group the x’s and y’s together...9x2 + 18x –4y2 + 16y = 43 Factor out the

GCF’s...9(x2 + 2x ) – 4(y2 – 4y ) = 43Complete the squares ...

+1 +4+ 9 – 16

9(x + 1)2 – 4(y – 2)2 = 36 36 36 36

(x +1)2 – (y – 2)2 = 1 4 9

a = 2 b = 3Center (–1, 2)