 # Hyperbola

A hyperbola is defined as an open curve having two branches which are mirror images to each other. Hyperbolic curves are of special importance in astronomy & space studies. Here, we will be studying about hyperbola and characteristics of such curves.

## What is Hyperbola?

A hyperbola is a locus of points in such a way that the distance to each focus is a constant greater than one. In other words, the locus of a point moving in a plane in such a way that the ratio of its distance from a fixed point (focus) to that from a fixed line (directrix) is a constant greater than 1. [Note: The point (focus) does not lie on the line (directrix)]

(PS/PM) = e > 1 eccentricity ### Standard Equation of Hyperbola

The equation of the hyperbola is simplest when the centre of the hyperbola is at the origin and the foci are either on the x-axis or on the y-axis.  The standard equation of a hyperbola is given as:

[(x2 / a2) – (y2 / b2)] = 1

where , b2 = a2 (e2  – 1)

### Important Terms and Formulas of Hyperbola

There are certain terms related to a hyperbola which needs to be thoroughly understood to be able to get confident with this concept. Some of the most important terms related to hyperbolae are:

• Eccentricity (e): e2 = 1 + (b2 / a2) = 1 + [(conjugate axis)2 / (transverse axis)2]
• Focii: S = (ae, 0) & S′ = (−ae, 0)
• Directrix: x=(a/e), x = (−a / e)
• Transverse axis:

The live segment A’A of length 2a in which the focii S’ and S both lie is called the transverse axis of the hyperbola.

• Conjugant axis:

The line segment B’B of length 2b between the 2 points B’ = (0, -b) & B = (0, b) is called the conjugate axis of the hyperbola.

• Principal axes:

The transverse axis & conjugate axis.

• Vertices:

A = (a, 0) & A’ = (-a, 0)

• Focal chord:

A chord which passes through a focus is called a focal chord.

• Double ordinate:

Chord perpendicular to the transverse axis is called a double ordinate.

• Latus Rectum:

Focal chord ⊥r to the transverse axis is called latus rectum.

Its length = (2b2 / a) = [(conjugate)2 / transverse] = 2a (e2 − 1)

The difference in focal distances is a constant

i.e. |PS−PS′| = 2a

Length of latus rectum = 2 e × (distance of focus from corresponding directrix)

End points of L.R : (± ae, ± b2 / a)

Centre:

The point which bisects every chord of the conic, drawn through it, is called the centre of the conic.

C: (0, 0) is the centre of [(x2 / a2) – (y2 / b2)] = 1

Note:

You will notice that the results for ellipse are also applicable for a hyperbola. You need to replace b2  by (−b2)

Recommended Video ### Practice Problems on Hyperbola

Example 1:

Find the equation of the hyperbola whose directrix is 2x + y = 1, focus (1, 2) and eccentricity √3.

Solution:

Let P(x, y) be any point on the hyperbola.

Draw PM perpendicular from P on the directrix,

Then by definition SP=ePM.

⇒ (SP)2 = e2 (PM)2 ⇒ (x − 1)2 + (y − 2)2 = 3{(2x + y – 1) / √(4+1)}2

⇒ 5 (x2 + y2 − 2x − 4y + 5)

= 3 (4x2 + y2 + 1 + 4xy − 2y − 4x)

⇒ 7x2 − 2y2 + 12xy − 2x + 14y – 22 = 0

Which is the required hyperbola.

Example 2:

Find the eccentricity of the hyperbola whose latus rectum is half of its transverse axis.

Solution:

Let the equation of hyperbola be [(x2 / a2) – (y2 / b2)] = 1

Then transverse axis = 2a and latus – rectum = (2b2 / a)

According to question (2b2 / a) = (1/2) × 2a

⇒ 2b2 = a2 (Since, b2 = a2 (e2 − 1))

⇒ 2a2 (e2 − 1) = a2

⇒ 2e2 – 2 = 1

⇒ e2 = (3 / 2)

∴ e = √(3/2)

Hence the required eccentricity is √(3/2)

## What is Conjugate Hyperbola?

2 hyperbolas such that transverse & conjugate axes of one hyperbola are respectively the conjugate & transverse axis of the other are called conjugate hyperbola of each other.

2 hyperbolas such that transverse & conjugate axes of one hyperbola are respectively the conjugate & transverse axis of the other are called conjugate hyperbola of each other.

(x2 / a2) – y2 b2 = 1 & (−x2 / a2) + (y2 / b2) = 1 are conjugate hyperbolas of each other.

(y2 / b2) – (x2 / a2) = 1

a2 = b2 (e2 − 1) Vertices: Vertices: (0,±b) L.R. = (2a2 / b) ### Some Important Conclusions on Conjugate Hyperbola

(a) If are eccentricities of the hyperbola & its conjugate, the

(1 / e12) + (1 / e22) = 1

(b) The foci of a hyperbola & its conjugate are concyclic & form the vertices of a square.

(c) 2 hyperbolas are similar if they have the same eccentricities.

(d) 2 similar hyperbolas are equal if they have the same latus rectum.

Example 3:

Find the lengths of transverse axis and conjugate axis, eccentricity, the co-ordinates of foci, vertices, length of the latus-rectum and equations of the directrices of the following hyperbola 16x2 − 9y2 = −144.

Solution:

The equation 16x2 − 9y2 = −144 can be written as (x2 / 9) – (y2 / 16) = −1

This is of the form (x2 / a2) – (y2 / b2) = −1

∴ a2 = 9, b2 = 16

⇒ a=3, b=4

Length of transverse axis: The length of transverse axis = 2b = 8

Length of conjugate axis: The length of conjugate axis = 2a = 6

Eccentricity: Foci: The co-ordinates of the foci are (0, ± be) i.e., (0, ± 5)

Vertices: The co-ordinates of the vertices are (0, ± b) i.e. (0, ± 4)

Lengths of latus-rectum: The length of latus-rectum = (2a2 / b) = [2(3)2] / 4 = 9/2

Equation of directrices: The equation of directrices are

y = ± (b/e) ⇒y = ± [4 / (5/4)] ⇒ y ± (16 / 5)

### Auxiliary Circles of the Hyperbola

A circle drawn with centre C & transverse axis as a diameter is called the auxiliary circle of the hyperbola. The auxilary circle of hyperbola equation is given as:

Equation of the auxiliary circle is x2 + y2 = a2,

Note from the following figure that P & Q are called the “corresponding points” of the hyperbola & the auxiliary circle. Parametric Representation:

x = a sec θ and y = b tan θ

i.e. If (a sec θ, b tan θ) is on the hyperbola, then Q : (a cos θ, a sin θ) lies on the auxiliary circle. The equation of chord joining 2 points P (α) and Q (f) is given by: The position of a point P

S1 = x12/a2 + y2/b2 = 1 is positive, zero or negative accordingly as (x1 ,y1) lies inside, on or outside is positive, zero or negative accordingly as (x1 ,y1) lies inside, on or outside.

Example 4: Find the position of the point (5, -4) relative to the hyperbola 9x2 – y2 = 1.

Solution:

Since 9(5)2 – (-4)2 – 1 = 225 – 16 – 1 = 208 > 0,

So, the point (5, -4) lies inside the hyperbola 9x2 – y2 = 1.

## Rectangular Hyperbola

The rectangular hyperbola is a hyperbola axes (or asymptotes) are perpendicular, or with its eccentricity is √2. Hyperbola with conjugate axis = transverse axis is a = b example of rectangular hyperbola.

x2/a2 – y2/b2

⇒ x2/a2 – y2/a2 = 1

Or, x2 – y2 = a2

We know b2 = a2 (e2 − 1) a2

= a2 (e2 − 1) e2 = 2e = √2. Eccentricity of rectangular hyperbola

Also, xy = c

e.g. xy = 1, y = 1/x

### Tangent of Rectangular hyperbola

The tangent of a rectangular hyperbola is a line that touches a point on the rectangular hyperbola’s curve. The equation and slope form of a rectangular hyperbola’s tangent is given as:

#### Equation of tangent

The y = mx + c write hyperbola x2/a2 – y2/b2 = 1 will be tangent if c2 = a2/m2 – b2.

#### Slope form of tangent

y = mx ± √(a2m2 – b2)

#### Secant

Secant will cut ellipse at 2 distinct points

⇒ c2 > a2 m2 – b2

#### Neither Secant Nor Tangent

For line to be neither secant nor tangent, quadratic equation will give imaginary solution.

⇒ c2 < a2 m2 – b2

Equation of tangent to hyperbola x2 /a2 – y2/b2 = 1 at point (x1 ,y1)

= (xx1)/a2 = (yy1)/b1 = 1

Parametric form of tangent:

(x secθ) /a − (y tanθ)/b = 1

Point of contact and examples on tangent

Compare:

y = mx + c

(xx1)/a2 – (yy1)/b2 = 1 – mx + y = c

x1 = (-a2c)/m;

y1 = -b2/c

(x1,y1) = [(-a2m)/c, -b2/c]

### Solved Examples on Hyperbola

Example 5: Show that the line x cos α + y sin α = p touches the hyperbola x2/a2 – y2/b2 = 1. If a2 cos2 α – b2 sin2 α = p2.

Solution:

Given line is x cos α + y sinα = p

y = − x cot α + p cosec α

Condition for y = mx + c to be tangent of hyperbola x2/a2 – y2/b2 = 1.

c2 = a2/m2 – b2

p2 cosec2α = α2 cot2α – b2

p2 = a2 cos2α – b2 sin2 α

Director circle:

Locus of point of intersection of ⊥r tangents.

Director circle for x2/a2 – y2 /b2 = 1 is

x2 + y2 = a2 – b2

Example 6: Find the equation of tangent to hyperbola x29−y2=1 whose slope is 5

Solution:

Slope of tangent m = 5, a2 = 9, b2 = 1

Equation of tangent in slope form is

y = mx ± √(a2m2 – b2 )

y = 5x ± √(9.52 – 1)

y = 5x ± 4√14

[Note: For ellipse, director circle is x2 + y2 = a2 + b2, x2/a2 + y2/b2 = 1]

Normal:

Equation of normal of x2/a2 – y2/b2 = 1 at (x1 ,y1)

a2x/x1 + b2y/y1 = a2 + b2

Normal in parametric form:

ax/secθ + by/tanθ = a2 + b2

Example 7: Find normal at the point (6, 3) to hyperbola x2/18 − y2/9 = 1

Solution:

Equation of Normal at point (x1, y1) is a2 = 18, b2 = 9

a2x/x1 + b2y/y1 = a2 + b2

Equation of Normal at point (6, 3) is

18x/6 + 9y/3 = 18 + 9

x + y = 9

Chord of contact:

T = 0

xx1/a2 − yy1/b2 = 1

Example 8: Find equation of chord of Contact of point (2, 3) to hyperbola x2/16 − y2/9 = 1

Solution:

Equation of chord of Contact is T = 0

i.e. (xx1)/a2 – (yy1)/b2 – 1 = 0

Or, 2x/16 – 3y/9 = 1

Or, x/8 – y/3 = 1

Equation of chord when mid-point is given

T = (xx1)/a2 – (yy1)/b2 – 1 = x12/a2 – y12/b2 − 1.

Example 9: Find the equation of chord of hyperbola x2/9 – y2/4 = 1 whose midpoint is (5, 1).

Solution:

Equation of chord of the hyperbola whose mid points is (5, 1)

T = (xx1)/a2 – (yy1)/b2 – 1 = x12/a2 – y12/b2 − 1

5x/9 – y/4 – 1= 25/9 – ¼ – 1

5x/9 – y/4 = 91/36