Conics

Find the equation of the parabola which has its vertex at the origin and its focus at point F in the following

cases.

a) F(2, 0)

b) F(0,-4)

c) F(-3,0)

d) F(0, 5)

In the Cartesian plane, represent the regions defined by the following relation.

x2 4y

Standing at the top of a cliff 32 m high, a person throws a stone which hits the ground 8 m from the base of

the cliff.

The following graph shows the trajectory of the stone. It is a part of a parabola whose vertex is at the origin.

What are the coordinates of the focus of this parabola?

1

2

3

The captain of a warship notices on his radar that a torpedo is approaching from the right. The head of the

torpedo is the shape of a parabola.

A cross section of the torpedo is shown in the Cartesian plane below.

Find the length (to the nearest centimetre) of the torpedo 50 cm from the tip of the head if the focus is 10 cm

from the vertex (tip).

A bridge is supported by a parabolic steel arch as shown in the following diagram.

S

Fhh

A B

Distance from the focus to the vertex : 8 m

Height : 12 m

Find the distance AB, which separates the feet of the arch.

10

10

y

x

4

5

The lampshade of a lamp is 30 cm in height. The top opening has an elliptical shape defined by the relation

1225400

22

yx

.

The bottom opening has an elliptical shape defined by the relation 1400625

22

yx

What is the length of the L, the slant height of the lampshade?

Find the standard form of the equation of the ellipse given its foci F1 and F2 and two vertices.

F1(-3, 0) , F2(3, 0) and V1(-5, 0) , V2(5, 0)

Find the standard form of the equation of the ellipse given its foci F1 and F2 and two vertices.

F1(0, -6) , F2(0, 6) and V1(0, -7) , V2(0, 7)

A flower garden in the shape of an ellipse is laid out by placing two posts 6 m apart and attaching the ends of a

10 m cord to each post.

What is the equation of the ellipse representing this flower bed if it is centered at the origin in the Cartesian

plane?

L

LAMPSHADE

Seen in perspectiveBased on the majoraxes of the ellipses

6

7

8

9

A factory manufactures windows that are in the shape of a semi-ellipse. The most popular model correspond

to the equation 116.0

22

yx

A metal strip is glued all along the top and the bottom of this window as shown in the diagram below.

A good approximation of the perimeter of an ellipse is found by using the formula

] b) +a b)(33 +(a b) +3(a [

where a and b represent respectively the length of each semi-axis in metres.

Rounded to the hundredth of a metre, what is the sum of the lengths of the two metal strips?

A)

4.00 m

C)

4.60 m

B)

4.14 m

D)

5.67 m

10

A railway tunnel goes through a mountain. A cross section shows that the tunnel is in the shape of the upper

half of an ellipse whose equation i

9x2 + 4y2 = 14

where the unit of measure is the metre.

The tunnel's lighting consists of lamps located at the

upper focus of the ellipse.

Rounded to the nearest hundredth, how far from the

ground are the lights?

A)

3.61 m

C)

5.33 m

B)

4.47 m

D)

7.21 m

The support for a swing is made of metal tubing in the shape of a parabola.

The vertex of the parabola is situated at point (0, 4) and the point at which the swing's chains are attached

coincides with the focus whose coordinates are (0, 3.85).

What is the equation of the parabola?

(0, 3.85)(0, 4)

y

x

11

12

A sculpture in the garden of a contemporary art museum consists of a circle and a parabola, as shown below.

The vertex of the parabola coincides with the centre O of the circle which has the equation

x2 + y2 32x 10y + 279 = 0 where the unit of measure is the metre.

The parabola, whose axis of symmetry is vertical, is constructed such that angle AOB is a right angle. Rounded

to the nearest hundredth, what distance is there between bases C and D of the sculpture?

Following the final event in the school's track and field meet, each member of the first place team will receive

a trophy. A cross section of the trophy is illustrated below.

The base of the trophy is the shape of a curve whose equation is x2 + 9y2 36 = 0 for y 0.

On top of this curve is a circle whose equation is x2 + y2 8y + 12 = 0.

The circle is topped by a section of a parabola.

The opening at the top of the trophy is level with the focus of the parabola. This opening is 18 cm wide.

What is the total height of the trophy?

A B

C D x

y

//

O

y (cm)

x (cm)

A

B

CD E

O

8 cm

13

14

The draft plan of the table Vivian designed is illustrated at the

right.

The curvature of the table leg corresponds to that of a

hyperbola whose equation is

324x2 56.25y2 = 18 225

where the unit of measure is the centimetre.

The line segment AB, passes through the focus F, and

represents the height of the table leg.

What is the height of the table leg?

In the adjacent diagram, the centre of the circle is

located at the upper focus of the ellipse defined

by :

1369

22

yx

The circle passes through the ends, A and B, of the

minor axis of the ellipse.

What is the degree measure of arc AB, situated inside the ellipse?

A

x

y

F

B

A B

F

0

y

x

15

16

A packaging company decorates its square boxes according to the model shown below.

The equation of the circle in the model has the form :

(x h)2 + (y k)2 = r 2

The equations of the parabolas are :

x2 8x y + 16 = 0 and x2 8x + y + 8 = 0

If the cartesian unit of measure is 1 cm, calculate the length of the diagonal of this box.

The equation of the hyperbola shown below is 4x2 y2 12 = 0.

Find the distance between foci F1 and F2.

F2 F1 x

y

17

18

A fire station is parabolic in shape. It is 24 metres wide at ground level and its maximum height is 9 metres.

Fire engines enter the station through three identical rectangular doors, each 4 metres high.

The front of the fire station is sketched below.

What is the area of the three doors, in square metres?

A company specializing in weight-lifting equipment is advertising a new type of barbell. A cross-cut view of the

barbell is sketched below. The radii of the two hemispherical ends measure 3 cm each. The centre part of the

barbell is defined by a hyperbola. Foci F1 and F2 lie on segments AD and CG respectively and the vertices are

4 cm apart.

What is the length of segment )MP (m ?

24 m

y

9

x

4

A B C

DE

F2 F1

G

M

P

x

y

Length

19

20

A car race takes place on an elliptical track. The foci of the ellipse are 500 m apart.

Luke and Nathalie, each located at one of the foci of the ellipse, were looking at the same car at a certain point

during the race. At that moment, each person's line of vision made an angle of 45 with the major axis.

Rounded to the nearest tenth, what is the shortest distance between Luke and the track?

The line whose equation is 4x + 3y 43 = 0 is tangent to a circle with centre C(3, 2).

What is the equation of this circle?

A geometric loci is defined by the set of points Q, whose sum of the distances from two fixed points M and N is

constant.

Which of the following equations represent this geometric loci?

A)

(x h)2 + (y k)2 = r2

C)

1)()(

2

2

2

2

= b

k y +

a

h x

B)

(y k)2 = 4c(x h)

D)

1)()(

2

2

2

2

- = b

k y

a

h x

Given the following:

1) The set of all points equidistant from a fixed point

2) The set of all points in which the product of the distances from two fixed points is constant

3) The set of all points such that each point in the set is equidistant from a line and a fixed point

Which of the statements above refer to a conic section?

A)

1 only

C)

1 and 3 only

B)

1 and 2 only

D)

2 and 3 only

22

23

24

25

The equations of two parabolas are given below:

y2 = -12x

(x 4)2 = 8(y 7)

The directrix of the first parabola intersects with the directrix of the second.

What are the coordinates of this point of intersection?

The logo of a graphics company is shown in the

diagram on the right. The logo is a rectangle

divided by a parabola.

In designing the logo, the company used the equation

y 2 = 50x

to draw the parabola.

The rectangle is labeled ABCD. Side AD is on the

directrix of the parabola and its focus is on side BC.

What is the length of side BC of the rectangle?

A)

25 units

C)

71 units

B)

50 units

D)

100 units

Focus

A B

CD

26

27

A carpenter places two nails on a rectangular piece of plywood as illustrated in the diagram below.

Each nail is placed 20 cm from each of the smaller edges. The smaller edges are 60 cm in length.

The carpenter uses these two nails as focal points to draw the largest ellipse possible on the rectangular plane.

Find the area of the rectangular piece of plywood.

A dome, in the shape of a semi-ellipse, protects a tennis court, as shown below.

20 m 3 m

8 m ?

The height of the dome at the centre is 8 m and its span is 20 m. Cameras must be fixed to the roof of the

dome at a horizontal distance of 3 meters from its edges.

At what height are the cameras from the ground?

.20 cm 20 cm60 cm

F1. F2

28

29

A student in Math 536 was designing a Halloween mask. He decided to draw an ellipse with equation

136100

22

yx

as illustrated below.

A B

C

He then drew two eyes, represented by circles with radii of 2 units. The centres of the circles are located directly above the foci of the ellipse. Next he drew a parabola which intersects two circles at points A and B respectively. Points A and B are on a line connecting the centres of the circles. The parabola is also tangent to a vertex of the ellipse at point C The final step was to draw a nose represented by the dark circle in the diagram. The centre of the circle is the focus of the parabola. What are the coordinates of the centre of the dark circle?

Consider a parabola with vertex (-1, 4) and focus (-4, 4).

Which of the following is the equation of the parabola in standard form?

A)

(y 4)2 = 12(x + 1)

C)

(y 4)2 = -12(x + 1)

B)

(x + 1)2 = 12(y 4)

D)

(x + 1)2 = -12(y 4)

30

31

Daphne wants to install a rectangular in-ground

pool in her backyard. She wants the pool,

measured in metres, to be inscribed within an

elliptical cement surface, as illustrated.

According to the design, the domain of the ellipse

would be [2, 10] and the range [6, 18].

The two ends of the pool pass through the foci of the ellipse.

What is the perimeter of Daphne's pool, to the nearest tenth of a metre?

Correction:

a) y2 = 8x b) x2 = -16y c) y2 = -12x d) x2 = 20y

32

1

The coordinates of the focus are

2

1- 0, .

2

3

Method : (example)

The equation of the parabola is y5 = 4ax. Since the focus is 10 cm from the vertex, a = 10; thus

y2 = 4ax = 4 10x = 40x

To find the width of the torpedo at 50 cm from the vertex, find the difference between the y intercepts at

x = 50.

y2 = 40x

y2 = 40 50

y2 = 2000

y1 = 2000 or y2 = - 2000

The width is equal to the absolute value of the difference between the y intercepts.

44.8972.4422000221 yy

Result : The width is 89 cm.

4

Work : (example)

The equation of a parabola with its vertex at the origin is x2 = 4cy. If the focus of the parabola is 8 m from the

vertex, c = 8 and

x2 = 4 8 y

To find the distance AB 12 m from the vertex, find the distance between the abscissas of the parabola for y =

12.

x2 = 4 8 12

x2 = 384

x1 = 384 or x2 = - 384

The distance AB is the absolute value of the difference between the abscissas.

2.3959.1923842ABm 21 xx

Answer : The distance AB is 39.2 m.

5

Work :

1.

In the representation of the lampshade based on the major axes

of the ellipses, consider triangle ABC, the half major axis ( OA )

of the upper ellipse and the half major axis ( .BO ) of the lower

ellipse.

2.

The measures of OA and .BO

Using 1225400

22

yx

, the measure of OA is found to be

20 cm.

Using 1400625

22

yx

, the measure of .BO is found to be

25 cm.

3.

The measure of BC

cm 5 = 20 25 = OA m BO' m = BC m

4.

Calculation of L (Using the Pythagorean relation)

925 = 25 + 900 = 5 + 30 = )BC (m + )AC (m = L22222

L 30.4 cm

Result : L 30.4 cm

L

6

1925

22

yx

14913

22

yx

11625

22

yx

C

B

7

8

9

10

11

The equation of the parabola is x2 = -0.6(y 4)

12

Work : (example)

Co-ordinates of O and measure of r

x2 + y2 32x 10y + 279 = 0

in canonic form

(x 16)2 + (y 5)2 = 2

thus O(16, 5) and r = 2

Measure of segment AB

)OB (m + )AO (m = AB m22

)2( + )2( = 22

2 = 4 = 2 + 2 =

Pythagorean theorem

Co-ordinates of point A

1 = AE m = OE m

x-co-ordinate of A = 16 1 = 15

y-co-ordinate of A = 4

In isosceles triangle AOB, which is right-

angled, the axis of symmetry of the parabola

divides segment AB, at the point of

intersection E, into two equal parts.

Point A belongs to the circle with centre O

whose equation is known.

Equation of the parabola

The vertex of the parabola is O(16, 5).

A B

C D x

y

//

O

E

13

y = a(x h)2 + k

4 = a(15 16)2 + 5

a = -1

y = -(x 16)2 + 5

The co-ordinates of point A are used to

calculate the value of a.

x-co-ordinate of points C and D

0 = -(x 16)2 + 5

x2 32x + 251 = 0

x1 = 16 5 and x2 = 16 + 5

Measure of segment CD

47.452516516CDm

Result : Rounded to the nearest hundredth, the distance between C and D is 4.47 m.

Note.- Also accept 4.48 m.

Work : (example)

Measure of segment OB

Transform the equation

x2 + y2 8y + 12 = 0

to the form

(x h)2 + (y k)2 = r2

which gives

x2 + (y 4)2 = 4.

Thus, r = 2 and the centre of the circle is (0, 4). The ordinate of B is 6.

Thus, m OB = 6 cm.

Measure of segment BC

The co-ordinates of B, the vertex of the parabola, are (0, 6). The co-ordinates of point E located

on the parabola are (4, 6 + c). Replacing h, k, x and y with their respective values in

(x h)2 = 4c(y k) gives (4 0)2 = 4c(6 + c 6).

Thus, c = 2

y (cm)

x (cm)

A

B

CD E

O

8 cm

14

and m BC = 2 cm

Height of the trophy

h = m OB + m BC

h = 6 + 2

h = 8 cm

Result : The total height of the trophy is 8 cm.

Work : (example)

Canonic form of the hyperbola is

22518

22518

22518

25,56

22518

324 22

yx

132426.56

22

yx

Coordinates of the focus F are

c2 = a2 + b2

c2 = 56.25 + 324

c2 = 380.25

c = ± 19.5

Therefore F(19.5, 0)

Solving for y

132425.56

5,19 22

y

1324

76.62

y

y2 = 324 5.76

y2 = 1866.24

A

x

y

F

B

15

y = ± 43.2

Height of the table leg

43.2 -43.2 = 86.4

Result : The height of the table leg is 86.4 cm.

Work : (example)

Equation of ellipse

1369

22

yx

Given this equation

3 = 9 =a (half-length of the minor axis of the ellipse)

6 = 36 = b (half-length of the major axis of the ellipse)

Measure of side FB = 6 units (half-length of major axis of the ellipse)

Since triangle FBO is a right triangle, then

30 = OFB mand2

1 =

6

3 =

FB m

OB m = OFBsin

Measure of angle AFB = 30 2 = 60 (y-axis being the axis of symmetry of AFB)

The measure of arc AB is 60 since angle AFB is at the centre.

A B

F

0

y

x

16

Result : The measure of arc AB is 60 degrees.

Work : (example)

Translating the two parabolas

x2 8x y + 16 = 0

x2 8x + 16 = y 16 + 16

(x 4)2 = y

t(4, 0)

x2 8x + y + 8 = 0

x2 8x + 16 = -y 8 + 16

(x 4)2 = -y + 8

(x 4)2 = -(y 8)

t(4, 8)

The origin of the Cartesian plane is located at the lower left corner of the drawing.

The centre of the circle is at (4, 4) and the diameter of the circle measures 8 units which corresponds to the

sides of the box.

The diagonal of the box will measure

8 + 8 = d 22

11.31d128d

Result : The length of the diagonal is 11.31 cm.

17

The distance between F1 and F2 is 7.7 units.

Accept any answer in [7.6; 8].

18

Example of an appropriate solution

Model

(x h)2 = 4c(y k)

(x 12)2 = 4c(y 9)

Value of c by substituting ordered pair (0, 0)

(0 12)2 =

144 =

-4 =

4c(0 9)

-36c

c

Equation of the parabola

(x 12)2 = -16(y 9)

Substituting y by 4 to find x1 and x2 :

(x 12)2 =

-16(4 9)

(x 12)2 =

80

x 12 =

80 or x 12 = - 80

x1

20.94 or x2 3.06

Measure of the base of the 3 doors

20.94 3.06 = 17.88

24 m

y

9

x

4

19

Area of the 3 doors

17.88 4 = 71.52 m2

Answer : The area of the three doors is 71.52 m2.

Accept an answer in [71, 72].

Example of an appropriate solution

Equation of the hyperbola

2

22

4 b

yx = 1

c = 22 ba

3 = 222 b

9 = 4 + b2

5 = b2

Therefore its equation is 154

22

yx

Ordinate of point C

54

3 22 y = 1

54

9 22 y = 1

5

- 2y =

4

9 1

y2 = 4

25 therefore y =

2

5

Length of segment MP

20

11 = 3 2 + 2

5 2 = MP m

Answer : The length of segment is 11 cm.

Example of an appropriate method

Diagram of the situation

Value of b

The distance c between each focus and the

centre of the ellipse is 250 m.

The triangle shown on the right is an isosceles

triangle. Hence, b = 250 m.

Value of a

a2 = b2 + c2

a2 = 2502 + 2502

a = 000125 353.553 m

Distance between Luke and the track

45 45

b

x

y

a

c

500 m

c = 250 m

45

b

x

y

22

a c = 000125 250 103.553 metres

Answer Rounded to the nearest tenth, the shortest distance between Luke and the track is

103.6 metres.

Example of an appropriate method

Equation of the tangent line

4x + 3y 43 = 0 or y = 3

43

3

4 x +

-

Equation of the perpendicular line passing through C(3, 2)

y = 4

1

4

3 x

Intersection of the two lines

3

43

3

4 x +

- =

4

1

4

3 x

12

175

12

25 x

x = 7

If x = 7, y = 54

17

4

3

23

The coordinate at the point of intersection is P(7, 5).

Measure of the radius of the circle

d(C, P) = 525253722

+

Equation of the circle (x 3)2 + (y 2)2 = 25

Answer The equation of the circle is (x 3)2 + (y 2)2 = 25.

The general form of the equation, x2 + y2 6x 4y 12 = 0, should also be accepted.

C

C

24

25

The point of intersection is (3, 5).

B

26

27

Example of an appropriate solution

Since width of rectangle is 60 cm,

minor axis measures 60 cm. So b = 30 cm.

If c represents distance between centre and

focal point, then c + 20 equals the length of

the semi-major axis, a.

Solving for c: a2 = b2 + c2

(c + 20)2 = 302 + c2

c2 + 40c + 400 = 900 + c2

40c = 500

c = 12.5

Therefore length of plywood is 2(c + 20) = 2(12.5 + 20) = 65

Area of plywood: 65 60 = 3900

Answer Rounded to the nearest cm2, the area of the plywood is 3900 cm2.

. .20 cm 20 cm60 cm

F1 F2

28

Example of an appropriate method

a = 10 b = 8

Find the equation of the semi-ellipse centre (0, 0) (Students may use other centres.)

164100

22

yx

Find the value of y at (7, y)

64100

49 2y = 1

64

2y =

100

491

y2 = 32.64

y +5.71 (-5.71 is rejected)

Answer: The cameras are 5.71 m from the ground.

Note: Do not penalize students who did not round their answers to the nearest hundredth.

Students who determined the equation of the semi-ellipse have shown a partial understanding of

the problem.

29

Example of an appropriate method

Determine location of foci of the ellipse:

c2 = a2 b2

c2 = 100 36

c2 = 64

c = ±8

Coordinates of foci at (±8, 0)

Since circles have a radius of 2 units,

coordinates of centres are (-8, 2) and (8, 2)

A B

C

Coordinates of points A and B are (-8 + 2, 2) and (8 2, 2) or (-6, 2) and (6, 2)

Coordinates of point C are (0, -6)

Equation of parabola in canonic form

(x h)2 = 4c(y k)

Substituting coordinates of points B and C into the equation to obtain value of c

B(x, y) = (6, 2) C(h, k) = (0, -6)

62 = 4c(2 -6)

36 = 4c(8)

36 = 32c

30

c = 8

9

32

36 or 1.125

Therefore coordinates of centre of dark circle =

8

74-,0

8

96-,0 or (0, -4.875)

Answer The coordinates of the centre of the dark circle are

8

74-,0 or (0, -4.875).

C

31

Example of an appropriate solution

Centre of ellipse (6, 12)

Equation of ellipse

1

36

12

16

622

yx

Foci

c2 = 36 16 = 20

c = 20 = 52 4,472...

Length of pool

2( 52 ) = 8.944 m

Coordinates of foci

(6, 12 ± 52 ) (6, 16.472), (6,7.528)

2 10

6

18

F1

Width (horizontal segment) of pool passes through, F1, (6, 16.472)

32

3

10,

3

26

3

86

9

646

256636

36

16

16

6

136

20

16

6

136

12472.16

16

6

2

2

2

2

22

xx

x

x

x

x

x

x

Width

m3

16

3

10

3

26

Perimeter

m28.6m555.285423

162

Answer: To the nearest tenth of a metre, the perimeter of the pool is 28.6 m.

Note: Do not penalize students who have not rounded their answer.

Students who have determined the length of the pool have shown they have a partial understanding of the problem.