Identifying Hyperbola and Its components, Lecture notes of Mathematics

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Lesson 1.4 Hyperbolas

Outcomes of the lesson:

At the end of the lesson, the student will be able to:

_1. Define a hyperbola

2. Determine the standard form and general for equation of a hyperbola
3. Graph a hyperbola in a rectangular coordinate system
4. Solve situational problems involving hyperbolas_

1.4.1 Definition, Components, and equation of a Hyperbola

Let 𝐹

1

and 𝐹

2

be two distinct points. The set of all points 𝑃, whose distances from 𝐹

1

and from 𝐹

2

differ by a

certain constant, is called a hyperbola. The points 𝐹

1

and 𝐹

2

are called the foci of the hyperbola.

The principal axis is the line that passes through the foci is called the Principal Axis. The points where the

hyperbola intersects the Principal Axis are called the Vertices (𝑽 𝟏

𝟐

). The points that is halfway between

the vertices is called the Center (𝑪). The segment joining the vertices is called the Transverse Axis (𝑽 𝟏

𝟐

). The

distance from the Center (𝑪) to any of the vertex is 𝒂. Thus the length of the transverse axis is 𝑽 𝟏

𝟐

= 𝟐𝒂. This

length is the constant difference referred to in the definition a hyperbola. The segment joining the points

𝟏

𝟐

is called the conjugate axis. The distance from the Center (𝑪) to any of the endpoints of the conjugate

axis 𝑬

𝟏

𝟐

is 𝒃. Hence, the length of the conjugate axis is 𝑬

𝟏

𝟐

= 𝟐𝒃. The foci 𝑭

𝟏

𝟐

are always located

on the principal axis. The distance from the Center (𝑪) to any of the foci 𝑭

𝟏

𝟐

is 𝒄 or 𝑭

𝟏

𝟐

= 𝟐𝒄. The two

diagonal dashed lines are the asymptotes of the hyperbola. These are lines where the branches of the graph of

the hyperbola approaches. It is used as a guide in graphing each branch of the hyperbola. The dashed rectangle is

called the Auxiliary Rectangle. It is also used as a guide in graphing. The dimension of this rectangle is 𝟐𝒂 𝒙 𝟐𝒃.

A hyperbola can have vertical or horizontal principal axis and the center can be at the origin or at any point in the

coordinate system.

Note: The branches in hyperbola are not parabolas.

1.4.2 Finding the center, the vertices, the endpoints of the conjugate axis, the foci, and the asymptotes of a Hyperbola

when the equation is given.

Example: For the hyperbola having the given equation, find the center, the vertices, the endpoints of the

conjugate axis, the foci, and the asymptotes. Sketch the graph.

𝑥

2

9

𝑦

2

16

2

2

Solution:

𝒙

𝟐

𝟗

𝒚

𝟐

𝟏𝟔

The given equation is of the form

𝒙

𝟐

𝒂

𝟐

𝒚

𝟐

𝒃

𝟐

= 𝟏 , thus the center is at the origin and the principal axis is

horizontal.

Solve for 𝒂 : Solve for 𝒃 :

𝒂

𝟐

= 𝟗 ⟹ 𝑎 = √ 9 ⟹ 𝒂 = 𝟑 𝒃

𝟐

= 𝟏𝟔 ⟹ 𝑏 = √ 16 ⟹ 𝒃 = 𝟒

Solve for 𝒄 :

𝒄 =

√ 𝒂

𝟐

• 𝒃

𝟐

⟹ 𝑐 = √ 9 + 16 ⟹ 𝑐 = √ 25 ⟹ 𝒄 = 𝟓

Center: 𝑪(𝟎, 𝟎)

Vertices: 𝑽

𝟏

( −𝒂, 𝟎

) ⟹ 𝑽

𝟏

( −𝟑, 𝟎

)

and 𝑽

𝟐

(𝒂, 𝟎) ⟹ 𝑽

𝟐

(𝟑, 𝟎)

Covertices: 𝑬

𝟏

( 𝟎, −𝒃

) ⟹ 𝑬

𝟏

( 𝟎, −𝟒

)

and 𝑬

𝟐

(𝟎, 𝒃) ⟹ 𝑬

𝟐

(𝟎, 𝟒)

Foci: 𝑭

𝟏

(−𝒄, 𝟎) ⟹ 𝑭

𝟏

(−𝟓, 𝟎)

and 𝑭

𝟐

( 𝒄, 𝟎

) ⟹ 𝑭

𝟐

( 𝟓, 𝟎

)

Asymptotes: 𝒚 = −

𝒃

𝒂

𝒙 ⟹ 𝒚 = −

𝟒

𝟑

𝒙

𝑎𝑛𝑑 𝒚 =

𝒃

𝒂

𝒙 ⟹ 𝒚 =

𝟒

𝟑

𝒙

𝟐

𝟐

Transform into standard form:

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

This equation is of the form

( 𝒚−𝒌

)

𝟐

𝒂

𝟐

( 𝒙−𝒉

)

𝟐

𝒃

𝟐

= 𝟏 , thus the principal axis is vertical with center at 𝐶(ℎ, 𝑘) =

Solve for 𝒂 : Solve for 𝒃 :

𝒂

𝟐

= 𝟗 ⟹ 𝑎 = √ 9 ⟹ 𝒂 = 𝟑 𝒃

𝟐

= 𝟒 ⟹ 𝑏 = √ 4 ⟹ 𝒃 = 𝟐

Solve for 𝒄 :

𝒄 = √𝒂

𝟐

• 𝒃

𝟐

⟹ 𝑐 = √

9 + 4 ⟹ 𝒄 = √

𝟏𝟑

Center: 𝑪(𝒉, 𝒌) ⟹ 𝑪(𝟏, −𝟐)

Vertices: 𝑽

𝟏

(𝒉, 𝒌 − 𝒂) ⟹ 𝑉

1

( 1 , − 2 − 3 ) ⟹ 𝑉

1

(𝟏, −𝟓)

and 𝑽

𝟐

( 𝒉, 𝒌 + 𝒂

) ⟹ 𝑉

2

( 1 , − 2 + 3

) ⟹ 𝑉

2

( 𝟏, 𝟏

)

Covertices: 𝑬

𝟏

(𝒉 − 𝒃, 𝒌) ⟹ 𝑬

𝟏

(𝟏 − 𝟐, −𝟐) ⟹ 𝑬

𝟏

(−𝟏, −𝟐)

and 𝑬

𝟐

(𝒉 + 𝒃, 𝒌) ⟹ 𝑬

𝟐

(𝟏 + 𝟐, −𝟐) ⟹ 𝑬

𝟐

(𝟑, −𝟐)

Foci: 𝑭

𝟏

(𝒉, 𝒌 − 𝒄) ⟹ 𝑭

𝟏

(𝟏, −𝟐 − √𝟏𝟑)

and 𝑭

𝟐

(𝒉, 𝒌 + 𝒄) ⟹ 𝑭

𝟐

(𝟏, −𝟐 + √

𝟏𝟑)

Asymptotes: 𝒚 − 𝒌 = −

𝒂

𝒃

(𝒙 − 𝒉)

𝑦 − (− 2 ) = −

3

2

(𝑥 − 1 )

𝒚 + 𝟐 = −

𝟑

𝟐

(𝒙 − 𝟏)

𝑎𝑛𝑑 𝒚 − 𝒌 =

𝒂

𝒃

( 𝒙 − 𝒉

)

𝑦 − (− 2 ) =

3

2

(𝑥 − 1 )

𝒚 + 𝟐 =

𝟑

𝟐

(𝒙 − 𝟏)

𝟏

and 𝑭

𝟐

, such that for any point on it, the absolute value of the di ff erence of its

distances from the foci is 𝟏𝟎.

The foci are horizontally aligned (the same 𝑦 − 𝑒𝑙𝑒𝑚𝑒𝑛𝑡). Thus, the hyperbola has horizontal principal

axis.

Find the center 𝑪

The midpoint of the foci is the center of the hyperbola.

Thus,

1

2

1

2

Find 𝒄 :

𝑐 is the distance from the center to the focus.

Since the center and the focus have same 𝑦 − 𝑣𝑎𝑙𝑢𝑒.

Then, the distance is determined by the absolute value of the difference between the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠.

Now,

𝟏

𝟏

𝟐

𝟐

. So, 𝑐 =

𝟐

𝟏

Find 𝒂 :

The given difference is 10 and this is the length of transverse axis ( 2 𝑎).

Therefore, 2 𝑎 = 10 ⟹

2 𝑎

2

10

2

Find 𝒄:

𝟐

𝟐

2

2

Find the standard equation:

Since the hyperbola has horizontal principal axis,

then,

(𝒙−𝒉)

𝟐

𝒂

𝟐

(𝒚−𝒌)

𝟐

𝒃

𝟐

Substituting,

(𝒙 − 𝒉)

𝟐

𝒂

𝟐

−

(𝒚 − 𝒌)

𝟐

𝒃

𝟐

= 𝟏

(𝑥 − 2 )

2

5

2

−

(𝑦 − (− 3 ))

2

( √

24 )

2

= 1

(𝒙 − 𝟐)

𝟐

𝟐𝟓

−

(𝒚 + 𝟑)

𝟐

𝟐𝟒

= 𝟏

𝟏

and 𝑽

𝟐

𝟏

and 𝑬

𝟐

The vertices have the same 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠. This means that the hyperbola has vertical principal axis.

Find the center 𝑪(𝒉, 𝒌):

The midpoint of the the segment joining the vertices 𝑉

1

(− 5 , − 3 ) and 𝑉

2

(− 5 , − 1 ), is the center of the

hyperbola.

Thus,

1

2

1

2

Find 𝒂 :

𝒂 is the distance from the center to the vertex.

Since the center and the vertices have same 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠.

Then, the distance is determined by the absolute value of the difference between the 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠.

Now,

𝟏

𝟏

𝟐

𝟐

So, 𝒂 =

𝟐

𝟏

Find 𝒃 :

𝑏 is the distance from the center to the endpoint of the conjugate axis.

Since the center and the endpoint of the conjugate axis have same 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠.

Then, the distance is determined by the absolute value of the difference between the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠.

Now,

𝟏

𝟏

𝟐

𝟐

So, 𝒃 =

𝟐

𝟏

Find the standard equation:

Since the hyperbola has vertical principal axis,

then,

(𝒚−𝒌)

𝟐

𝒂

𝟐

(𝒙−𝒉)

𝟐

𝒃

𝟐

= 𝟏. Substituting,

(𝒚 − 𝒌)

𝟐

𝒂

𝟐

−

(𝒙 − 𝒉)

𝟐

𝒃

𝟐

= 𝟏

(𝑦 − (− 2 ))

2

1

2

−

(𝑥 − (− 5 ))

2

2

2

= 1

( 𝒚 + 𝟐

)

𝟐

𝟏

−

( 𝒙 + 𝟓

)

𝟐

𝟒

= 𝟏