The eccentricity of the conjugate hyperbola to the hyperbola x^2-3y^2=1 is

Описание: 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.

(PS/PM) = e
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 hyperbola 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.

Conjugate 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)
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.

(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)
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
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.