Asymptotes are lines whose distance from the curve at infinity tends to zero Let y = mx + c is asymptote of H 
= – 1. Solving the two equations, we have (b 2 – a 2 m 2 ) x 2 – 2a 2 mcx – a 2 (b 2 + c 2 ) = 0. Both roots of this equation must be infinite so m = ± 
and c = 0 which implies that y = 
±x are asymptotes of 
= 1. Note that no real tangent can be drawn to the hyperbola from its centre and only one real tangent can be drawn from a point lying on its asymptote other than centre. Further combined equation of asymptotes is A = 
= 0 and conjugate hyperbola C = 
+ 1 = 0 . Hence 2A = H + C, as we can see, equation of A, H and C vary only by a constant, for asymptotes which can be evaluated by applying condition of pair of lines.
(i) The points of contact of tangents drawn to the hyperbola 
= 1 from point (2, 1) are
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(i) Equation of tangent is y = mx ± 
where a 2 m 2 – b 2 > 0 ⇒ y = mx ± 
As it passes through (2, 1) ⇒ 1 = 2m ± 
⇒ 1 – 2m = ± 
⇒ m = 1, 3
so tangents are y = x – 1 and y = 3x – 5
The corresponding points of contact are (3, 2) and 
(ii) Equation of tangent is y – 2 = m(x – 1) ⇒ y = mx + (2 – m)
it will be tangent if c 2 = a 2 m 2 – b 2 provided c ≠ 0
⇒ (2 – m) 2 = m 2 – 4 ⇒ m = 2 or ∞ ⇒ m = ∞ ⇒ tangent is x = 1
(iii) (1, 2) is centre of hyperbola so no real tangent can be drawn from it to hyperbola.
(iv) Since equation of a hyperbola and its asymptotes differ in constant terms only,
∴ Pair of asymptotes is given by xy – 3y – 2x + λ = 0
where λ is any constant such that it represents two straight lines.
∴ abc + 2fgh – af 2 – bg 2 – ch 2 = 0
⇒ 0 + 2 × – 
× – 1 × 
– 0 – 0 – λ 
= 0
∴ λ = 6
From (1), the asymptotes of given hyperbola are given by
xy – 3y – 2x + 6 = 0 or (y – 2) (x – 3) = 0
∴ Asymptotes are x – 3 = 0 and y – 2 = 0
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