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
