38 ellipses and hyperbolas

Conic Sections

Conic SectionsConic sections are the cross sections of right circular cones.

Conic SectionsConic sections are the cross sections of right circular cones.

A right circular cone

Conic SectionsConic sections are the cross sections of right circular cones. There are four different types of curves:


Conic SectionsConic sections are the cross sections of right circular cones. There are four different types of curves: • circles


Conic SectionsConic sections are the cross sections of right circular cones. There are four different types of curves: • circles • ellipses


Conic SectionsConic sections are the cross sections of right circular cones. There are four different types of curves: • circles • ellipses • parabolas


Conic SectionsConic sections are the cross sections of right circular cones. There are four different types of curves: • circles • ellipses • parabolas • hyperbolas A right circular cone

Conic SectionsConic sections are the cross sections of right circular cones. There are four different types of curves: • circles • ellipses • parabolas • hyperbolas

Where as straight lines are the graphs of 1st degree equations Ax + By = C, conic sections are the graphs of 2nd degree equations in x and y.



Where as straight lines are the graphs of 1st degree equations Ax + By = C, conic sections are the graphs of 2nd degree equations in x and y. In particular, the conic sections that are parallel to the axes (not tilted) have equations of the formAx2 + By2 + Cx + Dy = E, where A, B, C, D, and E are numbers.



Where as straight lines are the graphs of 1st degree equations Ax + By = C, conic sections are the graphs of 2nd degree equations in x and y. In particular, the conic sections that are parallel to the axes (not tilted) have equations of the formAx2 + By2 + Cx + Dy = E, where A, B, C, D, and E are numbers. We are to match these 2nd degree equations with the different conic sections.



Where as straight lines are the graphs of 1st degree equations Ax + By = C, conic sections are the graphs of 2nd degree equations in x and y. In particular, the conic sections that are parallel to the axes (not tilted) have equations of the formAx2 + By2 + Cx + Dy = E, where A, B, C, D, and E are numbers. We are to match these 2nd degree equations with the different conic sections. The algebraic technique that enable us to sort out which equation corresponds to which conic section is called "completing the square".



Where as straight lines are the graphs of 1st degree equations Ax + By = C, conic sections are the graphs of 2nd degree equations in x and y. In particular, the conic sections that are parallel to the axes (not tilted) have equations of the formAx2 + By2 + Cx + Dy = E, where A, B, C, D, and E are numbers. We are to match these 2nd degree equations with the different conic sections. The algebraic technique that enable us to sort out which equation corresponds to which conic section is called "completing the square".We start with the Distance Formula.


The Distance Formula:Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane, the distance r between P and Q is:

Conic Sections


r = (y2 – y1)2 + (x2 – x1)2

Conic Sections


r = (y2 – y1)2 + (x2 – x1)2 = Δy2 + Δx2

Conic Sections


r = (y2 – y1)2 + (x2 – x1)2 = Δy2 + Δx2 where

Conic Sections

Δy = the difference between the y's = y2 – y1


r = (y2 – y1)2 + (x2 – x1)2 = Δy2 + Δx2 where

Conic Sections


Δx = the difference between the x's = x2 – x1


r = (y2 – y1)2 + (x2 – x1)2 = Δy2 + Δx2 where

Example A. Find the distance between (2, –1) and (–2, 2).

Conic Sections




r = (y2 – y1)2 + (x2 – x1)2 = Δy2 + Δx2 where


Δy = (–1) – (2) = –3

Δy=-3

Conic Sections




r = (y2 – y1)2 + (x2 – x1)2 = Δy2 + Δx2 where


Δy = (–1) – (2) = –3 Δx = (2) – (–2) = 4

Δy=-3

Δx=4

Conic Sections




r = (y2 – y1)2 + (x2 – x1)2 = Δy2 + Δx2 where


Δy = (–1) – (2) = –3 Δx = (2) – (–2) = 4r = (–3)2 + 42 = 25 = 5

Δy=-3

Δx=4

r=5

Conic Sections




r = (y2 – y1)2 + (x2 – x1)2 = Δy2 + Δx2 where


Δy = (–1) – (2) = –3 Δx = (2) – (–2) = 4r = (–3)2 + 42 = 25 = 5

Δy=-3

Δx=4

r=5

Conic Sections

The geometric definition of all four types of conic sections aredistance relations between points.




r = (y2 – y1)2 + (x2 – x1)2 = Δy2 + Δx2 where


Δy = (–1) – (2) = –3 Δx = (2) – (–2) = 4r = (–3)2 + 42 = 25 = 5

Δy=-3

Δx=4

r=5

Conic Sections

The geometric definition of all four types of conic sections aredistance relations between points. We start with the circles.



CirclesA circle is the set of all the points that have equal distance r, called the radius, to a fixed point C which is called the center.

rr

CirclesA circle is the set of all the points that have equal distance r, called the radius, to a fixed point C which is called the center.

C

rr

The radius and the center completely determine the circle.

Circles

center

A circle is the set of all the points that have equal distance r, called the radius, to a fixed point C which is called the center.

r


Circles

Let (h, k) be the center of a circle and r be the radius.

(h, k)


r


Circles

(x, y)

Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center is r.

(h, k)


r


Circles

(x, y)

Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center is r. Hence,

(h, k)

r = (x – h)2 + (y – k)2


r


Circles

(x, y)


(h, k)

r = (x – h)2 + (y – k)2

orr2 = (x – h)2 + (y – k)2


r


Circles

(x, y)


(h, k)

r = (x – h)2 + (y – k)2

orr2 = (x – h)2 + (y – k)2

This is called the standard form of circles.


r


Circles

(x, y)


(h, k)

r = (x – h)2 + (y – k)2

orr2 = (x – h)2 + (y – k)2

This is called the standard form of circles. Given an equation of this form, we can easily identify the center and the radius.


r2 = (x – h)2 + (y – k)2

Circles

r2 = (x – h)2 + (y – k)2

must be “ – ”Circles

r2 = (x – h)2 + (y – k)2

r is the radius must be “ – ”Circles

r2 = (x – h)2 + (y – k)2

r is the radius must be “ – ”

(h, k) is the center

Circles

r2 = (x – h)2 + (y – k)2



Circles

Example B. Write the equation of the circle as shown.

r2 = (x – h)2 + (y – k)2



Circles


The center is (–1, 3) and the radius is 5. (–1, 3)

r2 = (x – h)2 + (y – k)2



Circles


The center is (–1, 3) and the radius is 5. Hence the equation is:52 = (x – (–1))2 + (y – 3)2

(–1, 3)

r2 = (x – h)2 + (y – k)2



Circles


The center is (–1, 3) and the radius is 5. Hence the equation is:52 = (x – (–1))2 + (y – 3)2 or25 = (x + 1)2 + (y – 3 )2

(–1, 3)

Example C. Identify the center and the radius of 16 = (x – 3)2 + (y + 2)2. Label the top, bottom, left and right most points. Graph it.

Circles


Put 16 = (x – 3)2 + (y + 2)2 into the standard form:

42 = (x – 3)2 + (y – (–2))2

Circles



42 = (x – 3)2 + (y – (–2))2

Hence r = 4, center = (3, –2)

Circles



42 = (x – 3)2 + (y – (–2))2


(3,-2)

Circles



42 = (x – 3)2 + (y – (–2))2


(3,-2)

Circles

When equations are not in the standard form, we have to rearrange them into the standard form. We do this by "completing the square".



42 = (x – 3)2 + (y – (–2))2


(3,-2)

Circles

When equations are not in the standard form, we have to rearrange them into the standard form. We do this by "completing the square". To complete the square means to add a number to an expression so the sum is a perfect square.



42 = (x – 3)2 + (y – (–2))2


(3,-2)

Circles

When equations are not in the standard form, we have to rearrange them into the standard form. We do this by "completing the square". To complete the square means to add a number to an expression so the sum is a perfect square. This procedure is the main technique in dealing with 2nd degree equations.

(Completeing the Square)Circles

(Completeing the Square)If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square,

Circles

(Completeing the Square)If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is the perfect square (x + b/2)2.

Circles


Circles

Example D. Fill in the blank to make a perfect square.

a. x2 – 6x + (–6/2)2


Circles


a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2


Circles


a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2

b. y2 + 12y + (12/2)2


Circles


a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2

b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2


Circles


a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2

b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2

The following are the steps in putting a 2nd degree equation into the standard form.


Circles


a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2

b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2

The following are the steps in putting a 2nd degree equation into the standard form.1. Group the x2 and the x-terms together, group the y2 and y terms together, and move the number term the the other side of the equation.


Circles


a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2

b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2

The following are the steps in putting a 2nd degree equation into the standard form.1. Group the x2 and the x-terms together, group the y2 and y terms together, and move the number term the the other side of the equation. 2. Complete the square for the x-terms and for the y-terms. Make sure add the necessary numbers to both sides.

Example E. Use completing the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom, left anf right most points. Graph it.

Circles


We use completeing the square to put the equation into the standard form:

Circles


We use completeing the square to put the equation into the standard form:x2 – 6x + + y2 + 12y + = –36

Circles


We use completeing the square to put the equation into the standard form:x2 – 6x + + y2 + 12y + = –36 ;complete squares x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36

Circles


We use completeing the square to put the equation into the standard form:x2 – 6x + + y2 + 12y + = –36 ;complete squares x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 ( x – 3 )2 + (y + 6)2 = 9

Circles


We use completeing the square to put the equation into the standard form:x2 – 6x + + y2 + 12y + = –36 ;complete squares x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 ( x – 3 )2 + (y + 6)2 = 9 ( x – 3 )2 + (y + 6)2 = 32

Circles


We use completeing the square to put the equation into the standard form:x2 – 6x + + y2 + 12y + = –36 ;complete squares x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 ( x – 3 )2 + (y + 6)2 = 9 ( x – 3 )2 + (y + 6)2 = 32 Hence the center is (3 , –6),and radius is 3.

Circles

Ellipses

EllipsesGiven two fixed points (called foci), an ellipse is the set of points whose sum of the distances to the foci is a constant.

F2F1


F2F1

P Q

R

If P, Q, and R are anypoints on a ellipse,


F2F1

P Q

R

p1

p2

If P, Q, and R are anypoints on a ellipse, thenp1 + p2


F2F1

P Q

R

p1

p2


= q1 + q2

q1

q2


F2F1

P Q

R

p1

p2


= q1 + q2

= r1 + r2

q1

q2

r2r1


F2F1

P Q

R

p1

p2


= q1 + q2

= r1 + r2

= a constant

q1

q2

r2r1


F2F1

P Q

R

p1

p2


= q1 + q2

= r1 + r2

= a constant

q1

q2

r2r1

Ellipses

An ellipse has a center (h, k );

(h, k)(h, k)

Given two fixed points (called foci), an ellipse is the set of points whose sum of the distances to the foci is a constant.

F2F1

P Q

R

p1

p2


= q1 + q2

= r1 + r2

= a constant

q1

q2

r2r1

Ellipses

An ellipse has a center (h, k ); it has two axes, the major (long)

(h, k)(h, k)


Major axis

Major axis

F2F1

P Q

R

p1

p2


= q1 + q2

= r1 + r2

= a constant

q1

q2

r2r1

Ellipses

An ellipse has a center (h, k ); it has two axes, the major (long) and the minor (short) axes.

(h, k)Major axis

Minor axis

(h, k)

Major axis

Minor axis


These axes correspond to the important radii of the ellipse.Ellipses

These axes correspond to the important radii of the ellipse. From the center, the horizontal length is called the x-radius

Ellipses

x-radius

x-radius

y-radius

These axes correspond to the important radii of the ellipse. From the center, the horizontal length is called the x-radius and the vertical length the y-radius.

Ellipses

x-radius

x-radius

y-radius


Ellipses

x-radius

The general equation for ellipses is Ax2 + By2 + Cx + Dy = E where A and B are the same sign but different numbers.

x-radius

y-radiusy-radius


Ellipses

x-radius

The general equation for ellipses is Ax2 + By2 + Cx + Dy = E where A and B are the same sign but different numbers. Using completing the square, such equations may be transform to the standard form of ellipses below.

x-radius

y-radiusy-radius

(x – h)2 (y – k)2

a2 b2

Ellipses

+ = 1

The Standard Form (of Ellipses)

(x – h)2 (y – k)2

a2 b2

Ellipses

+ = 1 This has to be 1.


(x – h)2 (y – k)2

a2 b2

(h, k) is the center.

Ellipses



(x – h)2 (y – k)2

a2 b2

x-radius = a


Ellipses



(x – h)2 (y – k)2

a2 b2

x-radius = a y-radius = b


Ellipses



(x – h)2 (y – k)2

a2 b2



Ellipses


Example A. Find the center, major and minor axes. Draw and label the top, bottom, right and left most points. (x – 3)2 (y + 1)2

42 22+ = 1


(x – h)2 (y – k)2

a2 b2



Ellipses



42 22+ = 1

The center is (3, –1).


(x – h)2 (y – k)2

a2 b2



Ellipses



42 22+ = 1

The center is (3, –1).The x-radius is 4.


(x – h)2 (y – k)2

a2 b2



Ellipses



42 22+ = 1

The center is (3, –1).The x-radius is 4.The y-radius is 2.


(x – h)2 (y – k)2

a2 b2



Ellipses


(3, -1)


42 22+ = 1

The center is (3, –1).The x-radius is 4.The y-radius is 2.


(x – h)2 (y – k)2

a2 b2



Ellipses


(3, -1) (7, -1)


42 22+ = 1

The center is (3, –1).The x-radius is 4.The y-radius is 2. So the right point is (7, –1),


(x – h)2 (y – k)2

a2 b2



Ellipses


(3, -1) (7, -1)

(3, 1)


42 22+ = 1

The center is (3, –1).The x-radius is 4.The y-radius is 2. So the right point is (7, –1), the top point is (3, 1),


(x – h)2 (y – k)2

a2 b2



Ellipses


(3, -1) (7, -1)(-1, -1)

(3, -3)

(3, 1)


42 22+ = 1

The center is (3, –1).The x-radius is 4.The y-radius is 2. So the right point is (7, –1), the top point is (3, 1), the left and bottom points are (1, –1) and (3, –3).


(x – h)2 (y – k)2

a2 b2



Ellipses


(3, -1) (7, -1)(-1, -1)

(3, -3)

(3, 1)


42 22+ = 1

The center is (3, –1).The x-radius is 4.The y-radius is 2. So the right point is (7, –1), the top point is (3, 1), the left and bottom points are (–1, –1) and (3, –3).


Example B. Put 9x2 + 4y2 – 18x – 16y = 11 into the standard form. Find the center, major and minor axis.Draw and label the top, bottom, right, left most points.

Ellipses


Group the x’s and the y’s:

Ellipses


Group the x’s and the y’s:9x2 – 18x + 4y2 – 16y = 11

Ellipses


Group the x’s and the y’s:9x2 – 18x + 4y2 – 16y = 11 factor out the square-coefficients9(x2 – 2x ) + 4(y2 – 4y ) = 11

Ellipses


Group the x’s and the y’s:9x2 – 18x + 4y2 – 16y = 11 factor out the square-coefficients9(x2 – 2x ) + 4(y2 – 4y ) = 11 complete the square

Ellipses


Group the x’s and the y’s:9x2 – 18x + 4y2 – 16y = 11 factor out the square-coefficients9(x2 – 2x ) + 4(y2 – 4y ) = 11 complete the square9(x2 – 2x + 1 ) + 4(y2 – 4y + 4 ) = 11

Ellipses


Group the x’s and the y’s:9x2 – 18x + 4y2 – 16y = 11 factor out the square-coefficients9(x2 – 2x ) + 4(y2 – 4y ) = 11 complete the square9(x2 – 2x + 1 ) + 4(y2 – 4y + 4 ) = 11 +9

Ellipses


Group the x’s and the y’s:9x2 – 18x + 4y2 – 16y = 11 factor out the square-coefficients9(x2 – 2x ) + 4(y2 – 4y ) = 11 complete the square9(x2 – 2x + 1 ) + 4(y2 – 4y + 4 ) = 11 +9 +16

Ellipses


Group the x’s and the y’s:9x2 – 18x + 4y2 – 16y = 11 factor out the square-coefficients9(x2 – 2x ) + 4(y2 – 4y ) = 11 complete the square9(x2 – 2x + 1 ) + 4(y2 – 4y + 4 ) = 11 + 9 + 16 +9 +16

Ellipses



Ellipses

9(x – 1)2 + 4(y – 2)2 = 36

9(x – 1)2 4(y – 2)2

36 36



+ = 1

Ellipses

9(x – 1)2 + 4(y – 2)2 = 36 divide by 36 to get 1

9(x – 1)2 4(y – 2)2

36 4 36 9



+ = 1

Ellipses


9(x – 1)2 4(y – 2)2

36 4 36 9



+ = 1

(x – 1)2 (y – 2)2

22 32 + = 1

Ellipses


9(x – 1)2 4(y – 2)2

36 4 36 9



+ = 1

(x – 1)2 (y – 2)2

22 32 + = 1

Ellipses


Hence, Center: (1, 2), x-radius is 2, y-radius is 3.

9(x – 1)2 4(y – 2)2

36 4 36 9



+ = 1

(x – 1)2 (y – 2)2

22 32 + = 1

Ellipses


Hence, Center: (1, 2), x-radius is 2, y-radius is 3.

(-1, 2) (3, 2)

(1, 5)

(1, -1)

(1, 2)

Hyperbolas

HyperbolasJust as all the other conic sections, hyperbolas are defined by distance relations.

Hyperbolas

Given two fixed points, called foci, a hyperbola is the set of points whose difference of the distances to the foci is a constant.

Just as all the other conic sections, hyperbolas are defined by distance relations.

A

If A, B and C are points on a hyperbola as shown

Hyperbolas


B

C


A

a2

a1

If A, B and C are points on a hyperbola as shown then a1 – a2

Hyperbolas


B

C


A

a2

a1

b2

b1

If A, B and C are points on a hyperbola as shown then a1 – a2 = b1 – b2

Hyperbolas


B

C


A

a2

a1

b2

b1

If A, B and C are points on a hyperbola as shown then a1 – a2 = b1 – b2 = c2 – c1 = constant.

c1

c2

Hyperbolas


B

C


HyperbolasA hyperbola has a “center”,

HyperbolasA hyperbola has a “center”, and two straight lines that cradle the hyperbolas which are called asymptotes.

HyperbolasA hyperbola has a “center”, and two straight lines that cradle the hyperbolas which are called asymptotes. There are two vertices, one for each branch.

HyperbolasA hyperbola has a “center”, and two straight lines that cradle the hyperbolas which are called asymptotes. There are two vertices, one for each branch. The asymptotes are the diagonals of a box with the vertices of the hyperbola touching the box.

HyperbolasA hyperbola has a “center”, and two straight lines that cradle the hyperbolas which are called asymptotes. There are two vertices, one for each branch. The asymptotes are the diagonals of a box with the vertices of the hyperbola touching the box. The asymptotes are the diagonals of a box with the vertices of the hyperbola touching the box.

HyperbolasThe center-box is defined by the x-radius a, and y-radius b as shown.

ab

HyperbolasThe center-box is defined by the x-radius a, and y-radius b as shown. Hence, to graph a hyperbola, we find the center and the center-box first.

ab

HyperbolasThe center-box is defined by the x-radius a, and y-radius b as shown. Hence, to graph a hyperbola, we find the center and the center-box first. Draw the diagonals of the box which are the asymptotes.

ab

HyperbolasThe center-box is defined by the x-radius a, and y-radius b as shown. Hence, to graph a hyperbola, we find the center and the center-box first. Draw the diagonals of the box which are the asymptotes. Label the vertices and trace the hyperbola along the asympototes.

ab

HyperbolasThe center-box is defined by the x-radius a, and y-radius b as shown. Hence, to graph a hyperbola, we find the center and the center-box first. Draw the diagonals of the box which are the asymptotes. Label the vertices and trace the hyperbola along the asympototes.

ab

The location of the center, the x-radius a, and y-radius b may be obtained from the equation.

HyperbolasThe equations of hyperbolas have the form Ax2 + By2 + Cx + Dy = E where A and B are opposite signs.

HyperbolasThe equations of hyperbolas have the form Ax2 + By2 + Cx + Dy = E where A and B are opposite signs. By completing the square, they may be transformed to the standard forms below.

(x – h)2 (y – k)2

a2 b2


– = 1 (x – h)2(y – k)2

a2b2 – = 1

(x – h)2 (y – k)2

a2 b2



– = 1 (x – h)2(y – k)2

a2b2 – = 1

(x – h)2 (y – k)2

a2 b2

x-rad = a, y-rad = b



– = 1 (x – h)2(y – k)2

a2b2 – = 1

(x – h)2 (y – k)2

a2 b2




– = 1 (x – h)2(y – k)2

a2b2

y-rad = b, x-rad = a

– = 1

(x – h)2 (y – k)2

a2 b2




– = 1 (x – h)2(y – k)2

a2b2


– = 1

(h, k)

Open in the x direction

(x – h)2 (y – k)2

a2 b2




– = 1 (x – h)2(y – k)2

a2b2


– = 1

(h, k)

Open in the x direction

(h, k)

Open in the y direction

HyperbolasFollowing are the steps for graphing a hyperbola.

HyperbolasFollowing are the steps for graphing a hyperbola.1. Put the equation into the standard form.

HyperbolasFollowing are the steps for graphing a hyperbola.1. Put the equation into the standard form.2. Read off the center, the x-radius a, the y-radius b, and draw the center-box.

HyperbolasFollowing are the steps for graphing a hyperbola.1. Put the equation into the standard form.2. Read off the center, the x-radius a, the y-radius b, and draw the center-box. 3. Draw the diagonals of the box, which are the asymptotes.

HyperbolasFollowing are the steps for graphing a hyperbola.1. Put the equation into the standard form.2. Read off the center, the x-radius a, the y-radius b, and draw the center-box. 3. Draw the diagonals of the box, which are the asymptotes. 4. Determine the direction of the hyperbolas and label the vertices of the hyperbola.

HyperbolasFollowing are the steps for graphing a hyperbola.1. Put the equation into the standard form.2. Read off the center, the x-radius a, the y-radius b, and draw the center-box. 3. Draw the diagonals of the box, which are the asymptotes. 4. Determine the direction of the hyperbolas and label the vertices of the hyperbola. The vertices are the mid-points of the edges of the center-box.

HyperbolasFollowing are the steps for graphing a hyperbola.1. Put the equation into the standard form.2. Read off the center, the x-radius a, the y-radius b, and draw the center-box. 3. Draw the diagonals of the box, which are the asymptotes. 4. Determine the direction of the hyperbolas and label the vertices of the hyperbola. The vertices are the mid-points of the edges of the center-box.5. Trace the hyperbola along the asymptotes.

HyperbolasExample A. List the center, the x-radius, the y-radius.Draw the box, the asymptotes, and label the vertices. Trace the hyperbola.

(x – 3)2 (y + 1)2

42 22– = 1

Center: (3, -1)


(x – 3)2 (y + 1)2

42 22– = 1

Center: (3, -1)x-rad = 4y-rad = 2


(x – 3)2 (y + 1)2

42 22– = 1

Center: (3, -1)x-rad = 4y-rad = 2

Hyperbolas

(3, -1)4

2

Example A. List the center, the x-radius, the y-radius.Draw the box, the asymptotes, and label the vertices. Trace the hyperbola.

(x – 3)2 (y + 1)2

42 22– = 1

Center: (3, -1)x-rad = 4y-rad = 2The hyperbola opens left-rt


(x – 3)2 (y + 1)2

42 22– = 1

(3, -1)4

2

Center: (3, -1)x-rad = 4y-rad = 2The hyperbola opens left-rt and the vertices are (7, -1), (-1, -1) .


(x – 3)2 (y + 1)2

42 22– = 1

(3, -1)4

2


Hyperbolas

(3, -1)(7, -1)(-1, -1) 4

2


(x – 3)2 (y + 1)2

42 22– = 1


Hyperbolas

(3, -1)(7, -1)(-1, -1) 4

2


(x – 3)2 (y + 1)2

42 22– = 1

When we use completing the square to get to the standard form of the hyperbolas, because the signs, we add a number and subtract a number from both sides.

Example B. Put 4y2 – 9x2 – 18x – 16y = 29 into the standard form. Find the center, major and minor axis. Draw and label the top, bottom, right, left most points.

Hyperbolas

Example B. Put 4y2 – 9x2 – 18x – 16y = 29 into the standard form. Find the center, major and minor axis. Draw and label the top, bottom, right, left most points. Group the x’s and the y’s:

Hyperbolas

Example B. Put 4y2 – 9x2 – 18x – 16y = 29 into the standard form. Find the center, major and minor axis. Draw and label the top, bottom, right, left most points. Group the x’s and the y’s:4y2 – 16y – 9x2 – 18x = 29

Hyperbolas

Example B. Put 4y2 – 9x2 – 18x – 16y = 29 into the standard form. Find the center, major and minor axis. Draw and label the top, bottom, right, left most points. Group the x’s and the y’s:4y2 – 16y – 9x2 – 18x = 29 factor out the square-coefficients4(y2 – 4y ) – 9(x2 + 2x ) = 29

Hyperbolas

Example B. Put 4y2 – 9x2 – 18x – 16y = 29 into the standard form. Find the center, major and minor axis. Draw and label the top, bottom, right, left most points. Group the x’s and the y’s:4y2 – 16y – 9x2 – 18x = 29 factor out the square-coefficients4(y2 – 4y ) – 9(x2 + 2x ) = 29 complete the square

Hyperbolas

Example B. Put 4y2 – 9x2 – 18x – 16y = 29 into the standard form. Find the center, major and minor axis. Draw and label the top, bottom, right, left most points. Group the x’s and the y’s:4y2 – 16y – 9x2 – 18x = 29 factor out the square-coefficients4(y2 – 4y ) – 9(x2 + 2x ) = 29 complete the square4(y2 – 4y + 4 ) – 9(x2 + 2x ) = 29

Hyperbolas

Example B. Put 4y2 – 9x2 – 18x – 16y = 29 into the standard form. Find the center, major and minor axis. Draw and label the top, bottom, right, left most points. Group the x’s and the y’s:4y2 – 16y – 9x2 – 18x = 29 factor out the square-coefficients4(y2 – 4y ) – 9(x2 + 2x ) = 29 complete the square4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29

Hyperbolas

Example B. Put 4y2 – 9x2 – 18x – 16y = 29 into the standard form. Find the center, major and minor axis. Draw and label the top, bottom, right, left most points. Group the x’s and the y’s:4y2 – 16y – 9x2 – 18x = 29 factor out the square-coefficients4(y2 – 4y ) – 9(x2 + 2x ) = 29 complete the square4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 16

Hyperbolas

Example B. Put 4y2 – 9x2 – 18x – 16y = 29 into the standard form. Find the center, major and minor axis. Draw and label the top, bottom, right, left most points. Group the x’s and the y’s:4y2 – 16y – 9x2 – 18x = 29 factor out the square-coefficients4(y2 – 4y ) – 9(x2 + 2x ) = 29 complete the square4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 16 –9

Hyperbolas

Example B. Put 4y2 – 9x2 – 18x – 16y = 29 into the standard form. Find the center, major and minor axis. Draw and label the top, bottom, right, left most points. Group the x’s and the y’s:4y2 – 16y – 9x2 – 18x = 29 factor out the square-coefficients4(y2 – 4y ) – 9(x2 + 2x ) = 29 complete the square4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9 16 –9

Hyperbolas

4(y – 2)2 – 9(x + 1)2 = 36


Hyperbolas

4(y – 2)2 – 9(x + 1)2 = 36 divide by 36 to get 1


Hyperbolas

9(x + 1)24(y – 2)2

36 36


– = 1


Hyperbolas

9(x + 1)24(y – 2)2

36 36


– = 1


Hyperbolas

9 4

9(x + 1)24(y – 2)2

36 36


– = 1


Hyperbolas

(y – 2)2 (x + 1)2

32 22 – = 1

9 4

9(x + 1)24(y – 2)2

36 36


– = 1


Hyperbolas

(y – 2)2 (x + 1)2

32 22 – = 1

Center: (-1, 2),

9 4

9(x + 1)24(y – 2)2

36 36


– = 1


Hyperbolas

(y – 2)2 (x + 1)2

32 22 – = 1

Center: (-1, 2), x-rad = 2, y-rad = 3

9 4

9(x + 1)24(y – 2)2

36 36


– = 1


Hyperbolas

(y – 2)2 (x + 1)2

32 22 – = 1

Center: (-1, 2), x-rad = 2, y-rad = 3The hyperbola opens up and down.

9 4

(-1, 2)

HyperbolasCenter: (-1, 2), x-rad = 2, y-rad = 3

(-1, 2)

(-1, 5)

(-1, -1)

HyperbolasCenter: (-1, 2), x-rad = 2, y-rad = 3The hyperbola opens up and down.The vertices are (-1, -1) and (-1, 5).

38 ellipses and hyperbolas

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