Hyperbola - 5 · KnowledgeBin.org

CONJUGATE HYPERBOLA

The hyperbola whose transverse & conjugate axes are respectively the conjugate & transverse axes of the standard hyperbolais called the conjugate hyperbola of the standard given hyperbola. The conjugate hyperbola of is = - 1.

SOME RESULTS OF CONJUGATE HYPERBOLA

(a) Length of Transverse axis       = 2b
(b) Length of conjugate axis        = 2a
(c) Foci                                     = (0, ±be)
(d) Equation of Directrices           = y = ± b/e
(e) Eccentricity                           = e =

(f) Length of Latus Rectum =

(g) Parametric co-ordinates = (a tan

, b sec

)
(h) Equation of transverse axis     = x = 0
(i) Equation of conjugate axis       = y = 0

Illustration : 11.

Find the length of transverse axis, conjugate axis eccentricity, coordinates of foci, vertices, length of latus rectum & equation of directrices of the hyperbola 9x² - 4y² = - 36.

Ans: The hyperbola is
     9x² - 4y² = - 36

= - 1 i.e of the form

= - 1 where a = 2 & b = 3
Length of transverse axis = 2b = 6units
Length of conjugation axis = 2a = 4 units
Eccentricity =

Vertices = (0, ± b) = (0, ± 3)
Length of Latus Rectum =

units
Equation of directrices = y = ± b/e
= y = ±

= y = ±

RECTANGULAR HYPERBOLA DEFINITION

(1) A hyperbola whose asymptotes include aright angle is called a rectangular Hyperbola. THe general form of the equation of hyperbola is x² - y² = a² = b².

Why ?
The asysmptotes of a hyperbola is
y = ±

If these are at right angles than
m₁m₂ = - 1

= - 1

b² = a²
Hence the equation of ractangular hyperbola is

= 1

x² - y² = a²

(2) If the length of transverse & conjugate axes of any hyperbola are equal, it is called as rectangular hyperbola.

Why ?
IF a = b, then

= 1 become

= 1

x² - y² = a²

RECTANGULAR HYPERBOLA (xy = c²)

The equation of Rectangular Hyperbola is x² - y² = a²
i.e. (x + y)(x - y) = a²
We know that x + y =0 & x - y = 0 are at 45⁰ & 135⁰ to the x axis. Now if we can rotate the axes through

= - 45⁰ without changing the origin.
So, we can replace (x, y) by
(x cos

- y sin

, x sin

+ y cos

) i.e.

The equation x² - y² = a² becomes

= a²

xy =

Let

= c² = any positive constant
Therefore xy = c² is the another form of Rectangular hyperbola.

RESULTS OF RECTANGULAR HYPERBOLA

(a) The parametric coordinates are x = ct & y = c/t.

(b) Equation of chord joining t₁ & t₂ is
x + yt₁t₂ = c(t₁ + t₂)

(c) Equation of tangent at t is

+ yt = x.

(d) Point of intersection of tangents at 't₁' & 't₂' is

.

(e) Equation of normal at 't' is
xt³ - yt - ct⁴ + c = 0.

(f) Points of intersection of normals at 't₁' & 't₂' is

Illustration : 12.

Find the eccentricity of Rectangular Hyperbola.

Ans: For the hyperbola

= 1; c =

Therefore for

= 1 we have b = a
Hence eccentricity of Rectangular hyperbola is

Easy

1. For what value of k does

= 1 represents a hyperbola ?

Ans: We know that the standard form of hyperbola can be

= 1 or

= - 1.

CASE - I

= 1
As a² & b² are +ve quantities greater than zero simultaneously, therefore
3 +

> 0 & 6 -

> 0

> - 3 &

< 6
Hence common solution is
- 3 <

< 6

CASE II

= - 1 or

= 1
Here also - a² & - b² are -ve quantities.
Therefore
3 +

< 0 & also 6 -

< 0

< - 3 &

> 6
There is no common solution.
Hence the only solution is
- 3 <

< 6 or

(- 3, 6)

Q.2. Find the value of c if the ellipse

= 1 & x² - y² = c² cut at right angles ?

Ans: The curves cut at right angle means that the tangents at the point of intersection of two curves are at right angles. If m₁ is the slope of the tangent to ellipse & m₂ is the slope of the tangent to hyperbola, then
m₁ =

.......................... (i)
m₂ =

....................................... (ii)
using (i) & (ii)
m₁m₂ = - 1

= 1

But

= 1 as (x₁y₁) lies on hyperbola

x₁² = 9/2 ................................................ (iii)

y₁² = 4/2 ............................................... (iv)
Bur x₁² - y₁² = c² .......................................... (v)

Using (iii) & (iv) in (v) we get
c² =

c = ±

Q.3. Prove that if normal to the hyperbola xy = c² at any point t₁ meets the curve again at t₂, then, t₁³t₂ = - 1.

Ans: Equation of normal at point t₁ is
yt₁ - xt₁³ = c - ct₁⁴
Now if this normal again meets at 't₂', then (ct₂,

) must satisfy the normal

t₁ - ct₂t₁³ = c - ct₁⁴

ct₁³[t₂ - t₁] =

[t₁ - t₂]

ct₁³t₂[t₂ - t₁] + c(t₂ - t₁) = 0

c(t₂ - t₁)(t₁³ + 1) = 0

As t₂

t₁ & c > 0
t₁³t₂ + 1 = 0

t₁³t₂ = - 1

Q.4. If (a sec

, b tan

) & (a sec

, b tan

) are the end points of a focal chord of the hyperbola

= 1, then prove that tan

tan

.

Ans: The equation of chord joining the points (a sec

, b tan

) & (a sec

, b tan

) is

It passes through focus (ae, 0), then

Applying componendo & dividendo we get

Q.5. Find the locus of poles of normal chords of the rectyangular hyperbola xy = c².

Ans: Equation of normal at any t is
            xt³ - yt = c(t⁴ - 1) ........................................ (i)
Lert its ple be P(x₁, y₁) then the equation of polar is
    xy₁ + x₁y = 2c²..................................................... (ii)
Comparing the coefficients of equations (i) & (ii) as they represent same equation

t² = -

.............................................................. (iii)
&

.......................................................... (iv)
From (iii) & (iv)

(x₁² - y₁²)² = - 4c²x₁y₁

Hence locus of (x₁, y₁) is

(x - y)² + 4c²xy = 0

Q.6. Find the co-ordinates of the foci & equation of the directrices of rectangular hyperbola xy = c² ?

Ans: When the hyperbola x² - y² = a² converts into xy = c² by rotation by - 45⁰ then a² = 2c².
So, for

= 1
Coordinates of foci are (± ae, 0) = (±

a, 0)
Also directrices are x = ±