Hyperbola - 4

Illustration : 6.
Find the equation of pair of tangents drawm from a point P(1, 1) to the hyperbola   = 1.

Ans: Here   a² = 2   b² = 1
     S = - 1
     S₁ = - 1 = -
     T = - 1 = - y - 1
Hence equation is   SS₁ = T²

- 3(x² - 2y² - 2) = (x - 2y - 2)²
4x² - 2y² - 4x + 8y - 4xy - 2 = 0
2x² - y² - 2x + 4y - 2xy - 1 = 0   is the equation of pair of tangents.


CHORD OF CONTACT:
If the tangent from a point P(x’, y’) to the hyperbola touch the equation of chord of contact QR is

Why ?
Let Q = (x₁, y₁)   &   R = (x₂, y₂)
We know that PQ & Pr are tangents, hence equation of PQ is
     = 1 ......................................... (i)
equation of PR is
     = 1 ........................................ (ii)
Since both (i) & (ii) passes through P(x', y')
  = 1 ........................................ (iii)
&   = 1 ........................................ (iv)
By looking carefully at (iii) & (iv) we can say that (x₁, y₁) & (x₂, y₂) lies on
     = 1     i.e.      T = 0.


POLE AND POLAR:


Let P(x₁, y₁) be any point inside or outside the hyperbola the if any straight line drawn through P intersects the hyperbola at A & B. Then the locus of points of intersection of tangents to the hyperbola at A & B is called the polar of the given point P with respect to hyperbola & the point P is called the pole of the polar.

Why ?
Let P be (x₁, y₁) & hyperbola be . If tangents to the hyperbola at A & B meets at Q(h, k), then AB is the chord ofcontact with respect to Q(h,k).
  Equation of AB is
    
But P(x₁, y₁) lies on it, hence
     = 1
Hence locus of Q(h, k) is
    
This is required equation of polar with (x₁, y₁) as its pole.


Illustration : 7.
Find the pole of given line y = mx + c with respect to hyperbola .
Ans: Let the pole of line be (, ) then the polar with respect to hyperbola is
     = 1
But we have given the polar to be
     y = mx + c
Hence by comparing the coefficients we get
    
  &  
Hence pole is  


DEAMETER:

The locus of the middle points of any set of parallel chords of a hyperbola is called a diameter & the point where diameter intersects the hyperbola iscalled the vertex of diameter & equation of diameter is   y = .

Why ?
Let y = mx + c be a set of parallel chords with c as a variable.


By solving y = mx + c , we get
     = 1
(a²m² - b²)x² + 2mca²x + a²(b² + c²) = 0
This equation has roots x₁ & x₂
  x₁ + x₂ = ......................................... (i)
But (h, k) is the middle point of QR then,
     h = .................................................. (ii)
From (i) & (ii)
     h =
c =
We also know that (h, k) lieson y = mx + c
  k = mh + c
k = mh + - mh
k = .


CONJUGATE DIAMETERS

Two diameters are said to beconjugate when each bisects all chords parallel to the others.

Why ?
Let one set of parallell chords be
     y = mx + c .............................................. (i)
then its diameter is   y =
Let other set of parallel chords be
     y = m₁x + c
then its diameter is   y = ........................ (ii)
But slope of equation (i) & (ii) are same, therefore
     = m
mm₁ =
Therefore product of the slopes of two conjugate diameter is .


Illustration : 8.
Find the condition for the lines 2Ax² + 2Hxy + By² = 0   to be conjugate diameter of   = 1.
Ans: Let the slope of lines whose equation is 2Ax² + 2Hxy + By² = 0   be   m₁ & m₂ then we know that
     m₁m₂ = ........................................... (i)
We also know that if m₁ & m₂ are the slopes of two conjugate diameters then
     m₁m₂ = ........................................... (ii)
Hence by (i) & (ii)
    
10A = 3B   is the required condition.


DIRECTOR CIRCLE:

The locus of the point of intersection of the tangent to the hyperbola , which are perpendicular to each other is called director circle & the equation of director is
     x² + y² = a² - b²      (a > b )

Why ?
The equation of tangent to hyperbola with slope m is
     y = mx + ................................... (i)
& the perpendicular tangent with slope - is
     y = - x + ................................... (ii)
x + my = .................................... (iii)
Squaring & adding equations (i) & (iii) we get
     (y - mx)² + (x + my)² = a²m² - b² + a² - b²m²
(1 + m²)(x² + y²) = (1 + m²)(a² - b²)
(1 + m²)[x² + y² - (a² - b²)] = 0
As   1 + m² 0, therefore
      x² + y² = a² - b² is the equation of director circle of hyperbola.


Illustration : 9.

Find the radius of the director circle of the hyperbola   .
Ans: We know the equation of director circle of is
     x² + y² = a² - b²
Here   a² = 4   &   b² = 2
Therefore   x² + y² = 4 - 2 = 2
x² + y² = ()² = r²
Here r = , Hence the radiusof director circle is units.


ASYMPTOTES

An asymptote of a hyperbola or any curve is a straight line which touches the curve at infinity at two points. The equation of asymptotes is   y = ±

Why ?
Let y = mx + c be an asymptote to
Hence solving bot we get
     (a²m² - b²)x² + 2a²mcx + a²(b² + c²) = 0
Now thishas two solutions for x & the asymptote touches the hyperbola at therefore both the roots must be . By looking at the above equation the only way that can be a solution is to be zero. Therefore
     a²m² - b² = 0    &    2a²mc = 0
m = ±    &    c = 0
Substituting m & c in y = mx + c
We get
     y = ± x   is the required equation of asymptotes.


Illustration : 10.
Find the angle between asymptotes of hyperbola   = 1.

Ans: THe equation of asymptotes is
     y = ± = ±
Now we have to find angle between two lines
     y =   &   y = -
If m₁ & m₂ are slopes than angle between them is
    
    
Hence the acute angle between the asymptotes is  
& the obstuse angle is