Let the equation of the hyperbola be x2/a2 – y2/b2 = 1
The tangent at P(x1, y1) on it is –xx1/a2 – yy1/b2 = 1. But (x1, y1) lies on the hyperbola
Eliminating y1 from the above equations, we find that the equations of two tangents to the curve at the point with abscissa x1 are
Taking limits when x1 tends to infinity, we have the equations of the asymptotes as x/a ± y/b = 0.
(i) There are two asymptotes both passing through the centre and equal inclined to the axis of x the inclination being tan–1 (b/a).
(ii) The angle between two asymptotes is 2 tan–1 (b/a).
(iii) The difference between the second degree curve and pair of asymptotes is constant.
(iv) A hyperbola and its conjugate hyperbola have the same asymptotes.
(v) The equation of the hyperbola and that of its pair of asymptotes differ by a constant. For example, if S = 0 is the equation of the hyperbola, then the combined equation of the asymptotes is given by S + K = 0. The constant K is obtained from the condition that the equation S + K = 0 represents a pair of lines. Finally the equation of the corresponding conjugate hyperbola is S+ 2K = 0.
(vi) If b = a then x2/a2 – y2/b2 = 1 reduces to x2 – y2 = a2. The asymptotes of rectangular hyperbola x2 – y2 = a2 are y = ± x which are at right angles.
Solved Example 1:
Find the equation of the of the hyperbola whose asymptotes are x + 2y + 3 = 0 and 3x + 4y + 5 = 0 and which passes through the point
(1, -1). Find also the equation of the conjugate hyperbola.
Combined equation of asymptotes is (x + 2y + 3) (3x + 4y + 5) = 0
or 3x2 + 10xy + 8y2 + 14x + 22y + 15 = 0 ….(i)
Also we know that the equation of the hyperbola differs from that of asymptotes by a constant.
Let the equation of the equation of the hyperbola differs from that of asymptotes by a constant.
Let the equation of the hyperbola be
3x2 + 10xy + 8y2 + 14x + 22y + λ = 0 ….(ii)
Since it passes through (1, –1) then 3(1)2 + 10(1)(–1) + 8(–1)2 + 14(1) + 22 (–1) + λ = 0
=> 3 –10 + 8 + 14 – 22 + λ = 0 => λ = 7
From (2), equation of hyperbola is
3x2 + 10xy + 8y2 +14x + 22y +7 = 0 ….(iii)
But we know that equation of conjugate hyperbola.
= 2 (Combined equation of asymptotes) – (Equation of hyperbola)
=> 6x2 + 20xy + 16y2 + 28x + 44y + 30 – 3x2 – 10 xy – 8y2 – 14x – 22y – 7 = 0
or 3x2 + 10xy + 8y2 + 14x + 22y + 23 = 0
Solved Example 2:
A tangent to the parabola x2 = 4ay meets the hyperbola xy = c2 in two points ‘P’ and ‘Q’. Prove that the mid point of PQ lies on a parabola.
Let ( h,k) be the mid point of PQ
Equation of PQ T = S1
hy+kx/2 –c2 = hk – c2
kx + hy – 2hk = 0 ……(1)
any point on x2=4ay is ( t , t2/4a)
equation of tangent
tx = 4a(t2/4a+y)/2 ……………(2)
tx – 2ay = t2/2
Comparing (1) and (2)
k/t = h/2a = 2hk/t2/2
t = –2ak/h also
h/–2a = 4hk/(–2ak/h)2 => 4a2k2/h2 = –8ak
h2= -a/2 k => x2 = –ay/2 Which is a parabola
Solved Example 3:
The equation of the line passing through the centre of a rectangular hyperbola is x – y – 1 = 0. If one of its asymptotes is 3x – 4y - 6 = 0 , then find the equation of the other asymptote.
Solution: Let centre of hyperbola (α, β)
Both lines pass through centre
∴ α – β –1 = 0, 3α – 4β – 6 = 0
α = –2, β = –3
slope of given asymptotes = 3/4
slope of required asymptotes = -4/3
∴ Equation is (y + 3) = –4/3 ( x + 2)
4x + 3y + 17 = 0