math 54 lecture 6 - conic sections

Introduction Parabola Ellipse Hyperbola Exercises

Conic Sections

Institute of Mathematics, University of the Philippines Diliman

Mathematics 54–Elementary Analysis 2

Conic Sections 1/ 23


Introduction

A conic section is a plane figure formed when a right-circular cone is cut by a plane.Below are samples of conic sections.



Parabola

The set of all points in the plane whose distance from the fixed point (focus) equalsthe distance from the fixed line (directrix).



Equation of a Parabola

Consider the directix and the focus situated this way:

A point P = (x,y) is on the parabola ifdistance of P to F = distance of P to D√(x−p)2 + (y−0)2 = ∣∣x− (−p)

∣∣√(x−p)2 +y2 = ∣∣x+p

∣∣(x−p)2 +y2 = (x+p)2

x2 −2px+p2 +y2 = x2 +2px+p2

y2 = 4px




For p > 0,

Focus Directrix Equation Orientation

(p,0) x =−p y2 = 4px right

(−p,0) x = p y2 =−4px left

(0,p) y =−p x2 = 4py upward

(0,−p) y = p x2 =−4py downward

The point on the parabola closest to the diretrix is called the vertex of the parabola.In every cases consider above, the vertex is at the origin.




In general, a parabola with a vertical or a horizontal directrix can take either of theform

(y−k)2 =±4p(x−h) (x−h)2 =±4p(y−k)

In these general cases, the vertex is located at the point (h,k).




For (y−k)2 = 4p(x−h) with p > 0,

Focus Directrix(p+h,k) x =−p+h(−p+h,k) x = p+h

For (x−h)2 = 4p(y−k) with p > 0,

Focus Directrix(h,p+k) y =−p+k(h,−p+k) y = p+k




Example

Find the focus, directrix and vertex of the parabola being described by theequation

y = 2x2 −4x+3

Solution:Note that the above equation can be written as

2(x−1)2 = (y−1)

which can be further written as

4

(1

2

)(x−1)2 = (

y−1)

.

Hence, the focus is at (1, 32 ), the directrix is y = 1

2 and the vertex is at (1,1).



Ellipse

The set of all points in the plane whose distances from the fix points (foci sing.focus) sum up to a constant.



Equation of an Ellipse

Foci : F1 = (c,0) and F2 = (−c,0)

Vertices : V1 = (a,0) and V2 = (−a,0)

Endpoints of minor axis: (b,0) and (−b,0)

Point P = (x,y) is on the ellipse if

distance of P to F1 + distance of P to F2 = constant2a√(x+ c)2 +y2 +

√(x− c)2 +y2 = 2a




√(x+ c)2 +y2 = 2a−

√(x− c)2 +y2

x2 +2cx+ c2 +y2 = 4a2 −4a√

(x− c)2 +y2 +x2 −2cx+ c2 +y2

4a√

(x− c)2 +y2 = 4a2 −4cx

a2(x2 −2cx+ c2 +y2

)= a4 −2a2cx+ c2x2

a2x2 −2a2cx+a2c2 +a2y2 = a4 −2a2cx+ c2x2(a2 − c2

)x2 +a2y2 = a2

(a2 − c2

)x2

a2+ y2

a2 − c2= 1




x2

a2+ y2

a2 − c2= 1

Notice that b2 = a2 − c2.

Thus, we have

x2

a2+ y2

b2= 1.




Forx2

a2+ y2

b2= 1 we have

Foci Vertices

a > b (±p

a2 −b2,0) (±a,0)

a < b (0,±p

b2 −a2) (0,±b)

The ellipses considered above are those centered at the origin.




In general, we may consider an ellipse with either a vertical or a horizontal majoraxis centered somewhere aside from the origin. They take the form

(x−h)2

a2+ (y−k)2

b2= 1

Foci Vertices Major Axis

a > b (h±p

a2 −b2,k) (h±a,k) horizontal

a < b (h,k±p

b2 −a2) (h,k±b) vertical

In the cases considered above, the ellipse is centered at (h,k).




Example

Find the equation of the ellipse centered at (1,1) with foci (−1,1) and (3,1) andvertices (−2,1) and (4,1).

Solution:The major axis of our ellipse is horizontal. From the data above, (h,k) = (1,1), a = 3,c = 2 and b =p

5.

Hence, we have

(x−1)2

9+ (y−1)2

5= 1



Hyperbola

The set of all points in the plane whose distance from the two fixed points (foci)that have a constant difference.



Equation of a Hyperbola

Foci : F1 = (c,0) and F2 = (−c,0)

Vertices : V1 = (a,0) and V2 = (−a,0)

From the above figure and from the definition of the hyperbola we have√(x+ c)2 +y2 −

√(x− c)2 +y2 = constant2a




Simplifying, we get

x2

a2− y2

c2 −a2= 1

Note that a < c. Thus, c2 −a2 > 0. Let b2 = c2 −a2. Then

x2

a2− y2

b2= 1




If we solve for y in terms of x we get

y =±√

x2b2

a2−b2

Consider the lines y = ba x and y =− b

a x. Verify the following:

limx→+∞

±√

x2b2

a2−b2

±b

ax

= 1

limx→−∞

±√

x2b2

a2−b2

∓b

ax

= 1

Hence, the lines y =± ba x serve as asymptotes of the hyperbola.




In general, for

x2

a2− y2

b2= 1

we have

a > b

Foci Vertices

(±√

a2 +b2,0) (±a,0)

a < b

Foci Vertices

(0,±√

a2 +b2) (0,±b)




In general, we may consider a hyperbola with either a vertical or a horizontaltransverse axis centered somewhere aside from the origin, say at (h,k). They takethe form

(x−h)2

a2− (y−k)2

b2= 1

a > b

Foci Vertices

(h±√

a2 +b2,k) (h±a,k)

a < b

Foci Vertices

(h,k±√

a2 +b2) (h,k±b)




Example

Find the equation of the hyperbola with foci (−1,1) and (4,1) and vertices (0,1) and(3,1).

Solution:

The desired hyperbola is centered at ( 32 ,1) so h = 3

2 and k = 1.

The value of a is the distance from the center to a vertex, which in this case is 32 .

The value of c is the distance of a focus to the center which is inthis case is 52 .

Now, b2 = c2 +a2 = 164 = 4. Hence, the desired equation is

(x− 32 )2(

32

)2− (y−1)2

(2)2= 1



Exercises

1 Identify the following conic sections:

a. 2x2 −3y2 +4x+6y−1 = 0b. 2x2 +3y2 +16x−18y−53 = 0c. 9x+y2 +4y−5 = 0d. 4x2 −x = y2 +1e. 7y−y2 −x = 0

2 Sketch the graph of the following conic sections:

a. x2 −2x+y = 0b. 9x2 −18x+4y2 = 27c. y = 3x2 −6x+5

d.y2

9− (x+2)2 = 1

e.(x−3)2

25− y2 +4y+4

9= 1

3 Do as indicateda. Determine the equation of the parabola whose focus and vertex are the vertex

and focus, respectively of the parabola with equation x+4y2 −y = 0.b. Let M = 3. Determine the equations of the hyperbola and ellipse having (±2,1) as

the foci and M as the length of the conjugate axis and minor axis, respectively.


math 54 lecture 6 - conic sections

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