Ann

132

Chapter 4 The Circle

Section 5.1 Conic Sections

5.1 Conic Sections Early Greek mathematicians are credited with developing the geometric properties of conic sections. Conic sections, or simply conics, are sections formed when planes cut at various angles a right circular cone of two nappes. Three types of conics are formed, namely the parabola, the ellipse (with circle as its special case), and the hyperbola.

Parabola

Circle

Ellipse

Hyperbola

Fig. 5.1.1

In addition to these sections formed, the plane may pass through the vertex of the cone and determine a point (point-ellipse), coincident lines, or two intersecting lines. An intersection of each of these kinds is called a degenerate conic. There are numerous applications of conics. For instance, they are used to study the paths of orbiting planets and satellites, and trajectories of projectiles. They are important in optics and atomic physics. They are also found in the different architectural designs.

Definition 5.1.1 Conic Section A conic section is the locus of a point that moves in a plane so that the ratio of its distance from a fixed point to its distance from a fixed line is a positive constant.

From the definition of a conic section, any point on this curve satisfies the relation (see Fig. 5.1.2.)

PF = e PD

⇒

e=

PF PD

Section 5.2 Parabola

Chapter 5 The Parabola

133

• General Parts of a Conic • Focus − the fixed point F. 2. Directrix − the fixed line D. 3. Eccentricity − the positive constant ratio denoted by e. 1.

Conic sections may be classified according to the value of eccentricity as follows:

L

•

P

•

D

•

If e = 1, the conic is a parabola. If e < 1, the conic is an ellipse (as e → 0, the ellipse approaches to a circle). 3. If e > 1, the conic is a hyperbola. 1. 2.

F

•

R

Fig. 5.1.2

5.2 Parabola One of the members of conic sections is a parabola. This conic has many practical applications. The construction of a parabolic reflector mirror is obtained by revolving a parabola about its symmetry. Suspension bridges are supported by parabolic cables suspended from two or more towers. When an object is given an initial thrust and then moves under the influence of the force of gravity and eventually strikes the ground, the path of the object is a parabola. Definition 5.2.1 Parabola A parabola is the locus of a point that moves in a plane so that its distance from a fixed point is equal to its distance from a fixed line. Its eccentricity is one.

• Parts and Properties of a Parabola • The fixed point F is called the focus. The fixed line D is called the directrix. The point on the parabola which is halfway from the focus to the directrix is the vertex. 4. The axis of symmetry (axis of the parabola) is the line passing through the focus and perpendicular to the directrix. This axis divides the parabola into two equal branches. 5. A chord connecting two points of the parabola passing through the focus and perpendicular to the axis of symmetry is called the right chord or latus rectum (LR). 1. 2. 3.

y L D

P

•

•

V

• •

F(a, 0)

R• x = −a

Fig. 5.2.1

x

134

Chapter 5 The Parabola

Section 5.3 Equations of Parabola

• Important Lengths and Distances Involved in a Parabola • 1. 2.

a = distance from vertex to focus or from vertex to directrix 2a = distance from focus to an end of latus rectum or a directrix

3. 4a = length of latus rectum 4. e = 1

5.3 Equations of Parabolas Consider the parabola in Fig. 5.2.1. This parabola has a vertex at (0, 0), focus at (a, 0), and a moving point P(x, y). From the definition of a parabola, we have PF = PD ( x − a ) 2 + ( y − 0) 2 = x − (−a ) ( x − a) 2 + y 2 = x + a ( x − a) 2 + y 2 = ( x + a)2 x 2 − 2ax + a 2 = x 2 + 2ax + a 2 y 2 = 4ax which is one of the standard equations of parabolas. A complete list of the standard equations of parabolas is summarized in Formulas 5.3.1a and 5.3.1b.

Formula 5.3.1a Standard Equations of Parabolas The standard equation of a parabola with vertex at (0, 0), I. axis of symmetry on OX, and A. right opening is B. left opening is

y 2 = 4ax

y 2 = −4ax

II. axis of symmetry on OY, and A. upward opening is B. downward opening is

x 2 = 4ay

x 2 = −4ay

Formula 5.3.1b Standard Equations of Parabolas The standard equation of a parabola with vertex at (h, k), I. axis of symmetry parallel or identical to OX, and A. right opening is B. left opening is

( y − k )2 = 4a ( x − h)

( y − k )2 = −4a ( x − h)

II. axis of symmetry is parallel or identical to OY, and A. upward opening is B. downward opening is

( x − h)2 = 4a( y − k )

( x − h)2 = −4a ( y − k )

Section 5.3 Equations of Parabolas

Chapter 5 The Parabola

Formula 5.3.2 Equations of Directrices of Parabolas The equation of the directrix of a parabola that opens A. to the right is B. to the left is

x = h−a

x = h+a

C. upward is

D. downward is

y = k −a

y = k+a

Formula 5.3.3 General Equations of Parabolas The general equation of a parabola with i. vertical axis of symmetry is

E≠ 0

x 2 + Dx + Ey + F = 0 ,

A, E′ ≠ 0

or Ax 2 + D′x + E ′y + F ′ = 0 , ii. horizontal axis of symmetry is

D≠0

y 2 + Dx + Ey + F = 0 , or Cy 2 + D′x + E ′y + F ′ = 0 , y

C, D′ ≠ 0

y

y

y

x

x

x

x

Graphs of Parabolas with Vertex at (0, 0) Fig. 5.3.1a y

y

y

y

x x

x

Graphs of Parabolas with Vertex at (h, k)

x

135

136

Fig. 5.3.1b

Chapter 5 The Parabola

Section 5.3 Equations of Para-

bolas

Example 5.3.1The following equations of parabolas are each in general form. 1. x 2 − 8 y = 0

5. x 2 + 4 x − 2 y = 0

2. 4 y 2 − 5 x = 0

6. 2 x 2 − 6 x − 13 y + 5 = 0

3. 2 y 2 − 7 x − 6 = 0

7. 2 y 2 − 8 x − 2 y + 15 = 0

4. x 2 − 24 y − 36 = 0

8. y 2 − 36 x − 6 y + 27 = 0

Example 5.3.2The following equations of parabolas are each in standard form. 1.

y2 = 4x

2 ( x + 2) 5. y = − 25 81

2.

x 2 = −12 y

2 6. ( x − 12 ) = 100( y + 6)

3.

( y − 3) 2 = 48 x

2 7. ( y − 23 ) = −24( x − 12 )

4.

x 2 = −36( y + 1)

8. ( y + 200) 2 = 2500( x − 150)

Example 5.3.3Reduce each of the following equations of parabolas in standard form. 1. x 2 − 4 x − 16 y + 4 = 0

2. x 2 − 2 x + 12 y − 35 = 0

3. 4 y 2 + 96 x + 4 y + 97 = 0

Solution. We reduce each given equation of the parabola to its standard form as follows. 1.

Given equation of the parabola: x 2 − 4 x − 16 y + 4 = 0 x 2 − 4 x − 16 y + 4 = 0 x 2 − 4 x = 16 y − 4 x 2 − 4 x + 4 = 16 y − 4 + 4 ( x − 2) 2 = 16 y , standard form 

2.

Given equation of the parabola: x 2 − 2 x + 12 y − 35 = 0 x 2 − 2 x + 12 y − 35 = 0 x 2 − 2 x = −12 y + 35 x 2 − 2 x + 1 = −12 y + 35 + 1 ( x − 2) 2 = −12 y + 36 ( x − 2) 2 = −12( y − 3) , standard form 

3.

Given equation of the parabola: 4 y 2 + 96 x + 4 y + 97 = 0 4 y 2 + 96 x + 4 y + 97 = 0 y 2 + 24 x + y + 974 = 0 y 2 + y + 14 = −24 x − 974 + ( y + 12 ) 2 = −24 x − 24

1 4

( y + 12 ) 2 = −24( x + 1) , standard form  Section 5.3 Equations of Parabolas

Chapter 5 The Parabola

137

Example 5.3.4 For each equation of the parabola, reduce it to its standard form and then find the direction of its opening, vertex, focus, and endpoints of the latus rectum. Determine the equation of the directrix and draw the parabola. 1.

y 2 − 16 x = 0

3. y 2 + 16 x − 32 = 0

2.

x 2 − 4 x + 8 y − 20 = 0

4. x 2 − 2 x − 24 y − 47 = 0

Solution. 1.

y 2 − 16 x = 0 ⇔ y 2 = 16 x takes the form y2 = 4ax. From this form, it follows that direction of opening: parabola opens to the right  4a = 16, length of the latus rectum, 2a = 8, distance from focus to an end of latus rectum, and a = 4, distance from vertex to focus (or to directrix).

Vertex: V(0, 0)  Latus rectum endpoints: L(4, 8), R(4, −8) 

Focus: F(4, 0)  Equation of directrix: x = −4 

y _ _ _ _ _ 6 _ _ 4 _ _ _ V | | | | | | _• −2 2 _ −2 _ _ _ −5 _ _ _ _ −9 _ 10

x = –4

|

| | −6

•

•

•

L(4, 8)

F(4, 0) | | | | 7

R(4, −8)

Fig. 5.3.2

2.

Reducing x 2 − 4 x + 8 y − 20 = 0 to its standard form, we have x 2 − 4 x = −8 y + 20

|

x

x 2 − 4 x + 4 = −8 y + 20 + 4 ( x − 2) 2 = −8( y − 3) 138

Chapter 5 The Parabola

Section 5.3 Equations of Parabolas

This takes the form ( x − h) 2 = −4a ( y − k ) from which it follows that direction of opening: parabola opens downward  −4a = −8 4a = 8, length of the latus rectum, 2a = 4, distance from focus to an end of latus rectum, and a = 2, distance from vertex to focus (or to directrix). Vertex: V(2, 3)  Latus rectum endpoints: L(−2, 1), R(6, 1) 

Focus: F(2, 1)  Equation of directrix: y = 5 

y

6 4 L(−2, 1) |

| | −6

|

|

•

_ _ _ y=5

_ _ V(2, 3) _ • _ F(2, 1)

•

| | | | | | _ −2 2 _ −2 _ _ _ −5

|

•

R(6, 1)

|

Fig. 5.3.3 3.

Reducing y 2 + 16 x − 32 = 0 to standard form, we have y 2 + 16 x − 32 = 0 y 2 = −16 x + 32 ( y − 0) 2 = −16( x − 2) This takes the form ( y − k ) 2 = −4a ( x − h) . Direction of opening: parabola opens to the left  –4a = –16 4a =16, length of the latus rectum 2a = 8, distance from focus to an end of latus rectum a = 4, distance from vertex to focus (or to directrix)

| | 7

|

x

Vertex: V(2, 0)  Latus rectum endpoints: L(–2, 8), R(–2, −8) 

Focus: F(–2, 0)  Equation of directrix: x = 6 

Section 5.3 Equations of Parabolas

Chapter 5 The Parabola

139

y

x=6 _ 10 _ _ L(–2, 8) _ • 6 _ _ 4 _ _ _ _ V(2, 0) F(–2, 0) | | | | | | | | | | | | | | | _ • −6 −3 • 7 _ −2 _ _ −5 _ _ _ R(–2, −8) _ • −9 _ _

|

x

Fig. 5.3.4

4.

Reducing x 2 − 2 x − 24 y − 47 = 0 to its standard form, we have x 2 − 2 x − 24 y − 47 = 0 x 2 − 2 x = 24 y + 47 x 2 − 2 x + 1 = 24 y + 47 + 1 x 2 − 2 x + 1 = 24 y + 48 ( x − 1) 2 = 24( y + 2) This takes the form ( x − h) 2 = 4a ( y − k ) so that the opening is upward. From this form, it follows

that 4a = 24, length of the latus rectum, 2a = 12, distance from focus to an end of latus rectum, and a = 6, distance from vertex to focus (or to directrix). Vertex: V(1, –2)  Latus rectum endpoints: L(−11, 4), R(13, 4) 

Focus: F(1, 4)  Equation of directrix: y = –8 

140

Chapter 5 The Parabola

Section 5.3 Equations of Parabolas

y

L(−11, 4)

_ _ 6 _ F(1, 4) _ _ • 3 _ _

•

|

|

|

–11

| | | | | –8 −6

|

|

•

| | | | | | | _ −2 2 _ • _ −3 V(1, –2) _ _ −5 _ _ –9

|

|

|

| | 9

|

| | 12

R(13, 4)

|

| 15

x

y = –8

_

Fig. 5.3.5

Exercise 5.1 A.

B.

For each equation of the parabola, find the direction of its opening, vertex, focus, and endpoints of the latus rectum. Determine the equation of the directrix and draw the parabola. 1. y 2 = −4 x

7. ( x − 3) 2 = −12( y − 4)

2. x 2 = 4 y

8. ( y + 2) 2 = 8( x − 1)

3. y 2 = −8( x − 2)

9. ( y − 1) 2 = −36( x − 1)

4. x 2 = −12( y + 1)

2 10. ( x + 12 ) = 16( y + 32 )

5. ( x + 4) 2 = 8 y

11. ( x + 5) 2 = −24( y − 4)

2 6. ( y − 12 ) = 20 x

12.

1 (x 12

+ 2) 2 = 3( y − 1)

For each equation of the parabola, reduce to its standard form and then find the direction of its opening, vertex, focus, and endpoints of the latus rectum. Determine the equation of the directrix and draw the parabola. 1. y 2 + 12 x = 0

7. x 2 + 2 x + 10 y − 19 = 0

2. x 2 − 2 x − 36 y + 1 = 0

8. y 2 − 12 x − 14 y + 25 = 0

3. y 2 + 24 x + 48 = 0

9. y 2 = −4(3 x + y + 4)

4. y 2 − 6 x − 4 y + 4 = 0

10. x = y 2 + 6 y + 12

5. x 2 − 4 x − 2 y + 2 = 0

2 11. y = − 14 x + x + 6

6. x 2 + 10 x − 8 y + 57 = 0

12. 4 y 2 − 16 y − 15 x + 16 = 0

Section 5.4 Parabolas Determined by Conditions

Chapter 5 The Parabola

141

5.4 Parabolas Determined by Conditions The equations of parabolas can be determined if enough conditions are given. Equations listed in Formulas 5.3.1 and 5.3.3 can be used to find such equations desired. We illustrate these in the following examples. Example 5.4.1Write the equation of the parabola with vertex at (0, 0) and focus at (2, 0). Solution. We plot the vertex and focus to determine the specific standard form of a parabola to be used (Fig. 5.4.1). We see from the illustration that y 2 = 4ax fits the conditions of the problem. We make the followy ing computations. _ 6

_ _ 4 _ _ _ F(2, 0) V | | | | | | | _• •2 −2 _ −2 _ _ _ −5 _

| VF | = a = | 2 − 0 | = 2 LR = 4a = 4(2) = 8 Hence, the equation of the parabola is y 2 = 8 x , standard form or

y 2 − 8 x = 0 , general form 

|

|

| | 7

|

x

Fig. 5.4.1

Example 5.4.2Write the equation of the parabola with vertex at (0, 0) and endpoints of the latus rectum at (–6, 3) and (6, 3). Solution. We plot the vertex and endpoints of latus rectum to determine the specific standard form of a parabola to be used (Fig. 5.4.2). We see from the illustration that x 2 = 4ay fits the conditions of the problem. We solve the length of the latus rectum to get the value of 4a. y LR = 4a = | 6 − (−6) | = 12 Hence, the equation of the parabola is

L(–6, 0)

•

x 2 = 12 y , standard form or

2

x − 12 y = 0 , general form 

|

| | –6

| |

_ _ 6 _ _ 4 _ _ _ | | | | | −2 V _• 2

Fig. 5.4.2

R(6, 0)

• |

|

| | 6

x

Example 5.4.3Write the equation of the parabola with vertex at (−2, 1) and equation of directrix y = 6. 142

Chapter 5 The Parabola

Section 5.4 Parabolas Determined by Conditions

Solution. The vertex is (−2, 1) so that h = −2 and k = 1. Since the directrix is horizontal and the vertex (− 2, 1) is below it, the parabola opens downward. (See Fig. 5.4.3.) Distance from vertex to directrix: a = |6 − 1| = 5 LR = 4a = 4(5) = 20 Thus, we have ( x − h ) 2 = −4 a ( y − k ) ( x + 2) 2 = −20( y − 1)

or x 2 + 4 x + 20 y − 16 = 0  y _ _ 5 _ _ 3 _ V(–2, 1) _

|

|

L

•

| | –11

|

|

|

| | –6

| |

y=6

•

| | | | | _ −2 2 _ –2 _ F _

•

|

|

| | 6

_ _ –6 _ _ –8

|

| | 9

•

R

|

x

Fig. 5.4.3

Example 5.4.4Find the equation of the parabola with focus (0, 6), axis on OY, and length of latus rectum 12. Solution. We first calculate the distance from vertex to focus. ⇒

LR = 4a = 12

a=3

Since the axis of parabola is on OY, the vertex is (0, k). There are two possible parabolas (Fig. 5.4.4); one that opens upward and the other one downward. Solving for k gives k1 = 6 − 3 = 3 First parabola : V1(0, 3); the opening is upward

or

k2 = 6 + 3 = 9

( x − h) 2 = 4a ( y − k )

⇒

( x − 0) 2 = 12( y − 3)

Section 5.4 Parabolas Determined by Conditions

Chapter 5 The Parabola

143

Simplifying the latter equation gives x 2 = 12( y − 3) , standard form or

x 2 − 12 y + 36 = 0 , general form 

Second parabola: V2(0, 9); the opening is downward ⇒

( x − h ) 2 = −4 a ( y − k )

( x − 0) 2 = −12( y − 9)

Simplifying the latter equation gives x 2 = −12( y − 9) , standard form or

x 2 + 12 y − 108 = 0 , general form  y _ _ 10 _ V2(0, 9) _ •_ 7_ _ 5_ _ •__ V (0, 3) 1 2 _

x 2 = −12( y − 9)

|

|

| | –11

|

|

|

| | –6

| |

| | | | | _ −2 2 –2 _ _ –4 _

x 2 = 12( y − 3)

|

|

| | 6

|

| | 9

|

x

Fig. 5.4.4

Example 5.4.5Find the equation of the parabola with axis horizontal and containing the points (9, 1), (9, –2), and (1, –1). Solution. The general equation of the parabola is useful in this type of problem. Since the axis is horizontal, we use y 2 + Dx + Ey + F = 0 . The three given points must satisfy this equation leading to a system of linear equations in D, E, and F. See Fig. 5.4.5. General Equation: y 2 + Dx + Ey + F = 0 (axis horizontal) At (9, 1) :

12 + D(9) + E (1) + F = 0 ⇔

9 D + E + F = −1

(1)

At (9, –2) : (−2) 2 + D(9) + E (−2) + F = 0 ⇔ At (1, –1) : 144

2

⇔

(−1) + D(1) + E (−1) + F = 0

9 D − 2 E + F = −4

(2)

D − E + F = −1

(3)

Chapter 5 The Parabola

Section 5.4 Parabolas Determined by Conditions

Let us solve these three linear equations simultaneously. (1) − (2)

(1) − (3)

9 D + E + F = −1 9 D − 2 E + F = −4 3E

At D = –1/4 and E = 1:

9 D − 2 E + F = −4 D − E + F = −1

D − E + F = −1 − ⇔ 14 − 1 + F = −1 F = 1/4

8D − E = –3 At E = 2: 8D – 1 = −3 D = –1/4

=3 E=1

Hence, the equation of the parabola is y 2 − 14 x + y + 14 = 0 ⇔ 4 y 2 − x + 4 y + 1 = 0 .  y 2_

(9, 1)

_

• _ –2 _

•

|

•

| 2

|

| 4

|

| 6

|

(1, –1)

| 8

|

| 10

x

•

(9, –2)

_ –4 _

Fig. 5.4.5

We can use the standard forms of the equations of parabolas as an alternative solution to this problem. The problem says that the axis of parabola is horizontal, hence we can use ( y − k ) 2 = 4a( x − h) or ( y − k ) 2 = −4a( x − h) , or equivalently ( y − k ) 2 = ±4a ( x − h) with the assumption that the opening is not known and then solve the value of ± 4a together with h and k. Standard form: ( y − k ) 2 = ±4a ( x − h) At (9, 1) : (1 − k ) 2 = ±4a (9 − h)

(1')

At (9, –2) : (−2 − k ) 2 = ±4a(9 − h)

(2')

At (1, –1) : (−1 − k ) 2 = ±4a (1 − h)

(3')

From equations (1') and (2'), we have (1 − k ) 2 = (−2 − k ) 2

1 1 − 2k + k 2 = 4 + 4k + k 2 ⇔ − 6k = 3 ⇔ k = − 2

⇔

Section 5.4 Parabolas Determined by Conditions

Chapter 5 The Parabola

From equations (1') and (3'), we can write

145

(1 − k ) 2 ( −1 − k ) 2 and = ±4 a = ±4a respectively and get 9−h 1− h

(1 − k ) 2 (−1 − k ) 2 . = 9−h 1− h But k = − 12 sot that (1 + 12 ) 2 9−h

=

(−1 + 12 ) 2 1− h

9 4

⇔

9−h

=

1 4

1− h

⇔

9 (1 − 4

h) = 14 (9 − h) ⇔

9 − 9h = 9 − h ⇔ h = 0 .

Solving for ± 4a : (−1 + 12 ) 2 = ±4a (1 − 0)

⇔

± 4a =

1 4

Note that the value of ±4 obtained is positive which means that the parabola opens to the right. Therefore, the equation of the parabola is ( y + 12 ) 2 = 14 ( x − 0) , standard form or

4 y 2 − x + 4 y + 1 = 0 , general form 

Exercise 5.2 A. Write the equation of the parabola with vertex at the origin and satisfying the given conditions. 1. Focus at (3, 0) 2. Focus at (–5/2, 0) 3. Focus at (0, –4) 4. Directrix is x – 5 = 0 5. Directrix is y + 4 = 0

6. Length of latus rectum is 25, opening is to the left 7. Focus on x-axis and containing the point (−3, 4) 8. Containing the point (3, –1) and opening is downward 9. Distance from focus to directrix is 36 and opening is upward 10. One endpoint of latus rectum is on y = 3, opening is upward

B. Write the equation of the parabola satisfying the given conditions. 1. 2. 3. 4. 5.

6. 7. 8.

Vertex (2, 1), focus (2, −1). Vertex (−1, 2), focus (0, 2). Vertex (2, 3), equation of directrix y − 6 = 0. Vertex (−2, 5), equation of directrix x + 4 = 0. Vertex (−3/2, 2), containing the point (−1, −1), latus rectum parallel to OX. Vertex (0, 2), axis parallel to OX, passing through (3, 8). Vertex (2, 1), axis parallel to OY, latus rectum 6. Vertex (−2, 4), latus rectum 8, opens to the left.

Vertex (−2, −2), latus rectum 10, opens upward Passing through (−7/2, 2), (−14/3, −5), axis parallel to OX, latus rectum 6. Passing through (2, 2), (2, −2), (−1, 1), axis horizontal. 12. Vertex on OX, latus rectum 8, containing (4, 4), axis on OX. 13. Vertex on the line x = −1, axis parallel to OX, containing the points (1, 0), (7, 3). 9. 10. 11.

146

Chapter 5 The Parabola

14. 15. 16. 17.

Section 5.4 Parabolas Determined by Conditions

Focus (1, −4), one end of latus rectum (1, −1). Axis parallel to OY, vertex on OX, containing the points (4, 2), (−2, 8). Axis parallel to OY, passing through (−1, −1), (0, 0), and (−3, 3). Axis horizontal, containing the points (0, −1), (0, 5), and (8, 1).

C. Find the points of intersection of the given curves. Sketch their graphs and shade the region enclosed. 1. x = 3 − y , y 2 − 4 x = 0 2. x − y − 3 = 0 , x 2 − 4 y − 12 = 0

4. x + 4 y − 8 = 0 , x 2 + 8 y − 16 = 0

3. 2 x − y − 12 = 0 , y 2 − 4 x = 0

6. y 2 − 2 y + 2 x + 5 = 0 , y 2 − 2 y + x + 1 = 0

5. x 2 + x − y = 0 , y 2 = −2 x

D. Sketch the region enclosed by the given set of curves. 1. 2.

y = x 2 , x 2 + y − 8 = 0 , 4 x − y + 12 = 0 4 x − y + 12 = 0 , 7 x + 2 y − 24 = 0 , x 2 + y − 9 = 0

E. Find the equation of the locus as indicated for the specified condition. Locus of a point equidistant from the point (4, 2) and the line y = −2. 2. Locus of a point equidistant from the point (2, −3) and the line x = 1. 2 2 3. Locus of the center of a circle tangent to the circle x + y − 9 = 0 externally and the line y = −5 . 1.

4.

Locus of the center of a circle tangent to the circle x 2 + y 2 − 2 x + 4 y + 1 = 0 externally and the x = −4 and. Find the locus of its center.

F. Solve as indicated. A circle touches the line x = 5 and the circle x 2 + y 2 − 4 = 0 . Find the locus of its center. (Two solutions). 2. Find the equation of the circle passing through the vertex and endpoints of the latus rectum of the parabola y 2 = 8 x . 1.

G.

Classify each equation of the curves as to a straight line (L), a circle (C), a parabola (P), or neither of the three (N). _____ a. x 2 + y 2 − 6 x + 4 y + 1 = 0 _____ b. x 2 − 2 x + 4 y + 1 = 0 _____ c. 2 x + 4 y + 1 = 0 _____ d. x 2 + 4 y 2 − 16 = 0

_____ i. y = −2 x 2 + 4 x + 1 _____ j. y = 4 x − 6 _____ k. 4 y 2 = −2( x + 4) _____ l. ( y − 4) 2 = 16(2 − x)

_____ e. y − 2 = −4( x − 5) x−2 y+2 + =1 3 9 _____ g. y 2 = −2 x + 4 y + 1

_____ m.

( x − 2) 2 ( y − 4) 2 + =1 16 16

_____ f.

_____ n. y 2 = x 2 + 4 y + 1

_____ h. 4 x 2 + 4 y 2 − 2 x + 4 y − 7 = 0

_____ o. y 2 = −2 x 2 + 8 y + 1 _____ p. y = −2