Math 54 Lecture 7 - Conic Sections (Hyperbola and Focus Directrix Equation)

Hyperbola Eccentricity Focus-Directrix Equation Exercises

Conic Sections(Hyperbola and Focus-Directrix Equation

)

Institute of Mathematics, University of the Philippines Diliman

Mathematics 54–Elementary Analysis 2

Hyperbola and Focus-Directrix Equation 1/ 19


Hyperbola

A hyperbola is a set of points in the plane whose distances from two fixed points(focuses/foci) 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 = 2a




Simplifying, we get

x2

a2− y2

c2 −a2= 1

Since c > a , we can let b2 = c2 −a2. Thus, obtaining

x2

a2− y2

b2= 1

Note that the rectangle formed above is called the auxiliary rectangle, and the redline segment, the conjugate axis of the hyperbola.




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, a hyperbola centered at the origin has form:

x2

a2− y2

b2= 1

Foci(c,0), (−c,0)

Vertices (horizontal transverse axis)

(a,0), (−a,0)

y2

b2− x2

a2= 1

Foci(0,c), (0,−c)

Vertices (vertical transverse axis)

(0,b), (0,−b)

Note: a2 +b2 = c2



Hyperbola

Example.

Sketch the graph of x2 −y2 = 4. Identify the foci.

Solution. The equation can be written asx2

22− y2

22= 1.

Hence, a = 2, b = 2, and transverse axis is horizontal, with center at (0,0).

vertices : V1(2,0), V2(−2,0)

draw the auxiliary rectangle

draw the asymptotes

c2 = a2 +b2= 8= (2p

2)2

foci : F1(2p

2,0), F2(−2p

2,0)



Hyperbola

Example.

Sketch the graph of 9y2 −4x2 = 144. Identify the foci.

Solution. The equation can be written asy2

42− x2

62= 1.

Hence, b = 4, a = 6, and transverse axis is vertical, with center at (0,0).

vertices : V1(0,4), V2(0,−4)

draw the auxiliary rectangle

draw the asymptotes

c2 = a2 +b2= 42 +62= 52=(2p

13)2

foci : F1(0,2p

13),F2(0,−2

p13)




In general, suppose the hyperbola is shifted so that the center is at the point (h,k).Then the form of the hyperbola is either

(x−h)2

a2− (y−k)2

b2= 1

Vertices Transverse Axis(h±a,k) horizontal

(y−k)2

b2− (x−h)2

a2= 1

Vertices Transverse Axis(h,k±b) vertical

Note: a2 +b2 = c2




Example

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

Solution:

Half-way between the foci (or bet. vertices) is the center,(

32 ,1

).

Note that a is the distance from the center to a vertex. Thus, a = 32 .

Also, c is the distance from the center to a focus. Thus, c = 52 .

Now, a2 +b2 = c2 =⇒ b2 = 254 − 9

4 = 164 = 4.

Since the transverse axis is horizontal, the equation is

4(x− 3

2

)2

9− (y−1)2

4= 1.



Hyperbola

Exercise.

Sketch the graph of the following hyperbolas:

1 x2 −y2 +6x−4x = 4

2 3y2 −4x2 −8x−24y−40 = 0



Eccentricity

For the parabola, we define the eccentricity e to be e = 1.For the ellipse and hyperbola, we define eccentricity to be

e = distance between foci

distance between vertices

Hence, we have

e = 0 circle

0 < e < 1 ellipse

e = 1 parabola

e > 1 hyperbola



Consider the ellipse(x−h)2

a2+ (y−k)2

b2= 1.

The lines perpendicular to the major axis of this ellipse at distances ± ae from the

center are the directrices of this ellipse.

Let D1 be the directrix nearest the focus F1 and let D2 be the directrix nearest thefocus F2. This pairing of the foci and the directrices will be refered to as thefocus-directrix pairing.



PF1 =√

(x+ c)2 +y2

=√

(x+ae)2 +b2 − b2

a2x2

=√

a2x2 +2a3ex+a4e2 +a2b2 −b2x2

a2

=√

c2x2 +2a3ex+a4e2 +a2b2

a2

=√

a2e2x2 +2a3ex+a4e2 +a2b2

a2

=√

e2x2 +2aex+a2e2 +b2

=√

e2x2 +2aex+ c2 +b2

=√

e2x2 +2aex+a2

= e

√x2 +2

a

ex+

( a

e

)2 = e

√(x−

(−a

e

))2 = e∣∣∣x− (

−a

e

)∣∣∣ = e ·PD1



Thus, if P is a point on the ellipse then

PF1 = e ·PD1 PF2 = e ·PD2



Consider the hyperbola(x−h)2

a2− (y−k)2

b2= 1.

The lines perpendicular to the transverse axis of this hyperbola at distances ± ae

from the center are the directrices of this ellipse.

Let D1 be the directrix nearest the focus F1 and let D2 be the directrix nearest thefocus F2. This pairing of the foci and the directrices will be refered to as thefocus-directrix pairing.



Similarly, if P is a point on the ellipse then

PF1 = e ·PD1 PF2 = e ·PD2



Focus-Directrix Equation

Hence, for any focus-directrix pair in an ellipse, hyperbola or parabola we have thefollowing equation

PF = e ·PD

this equation is refered to as the focus-directrix equation.



Exercises

1 Identify the following conic sections and determine their eccentricity:

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 Determine the equation of the parabola whose focus and vertex are the vertexand focus, respectively of the parabola with equation x+4y2 −y = 0.

3 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.


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Math 54 Lecture 7 - Conic Sections (Hyperbola and Focus Directrix Equation)