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Examples on Propositions of Hyperbola

 

Illustration:

Find the hyperbola whose asymptotes are 2x – y = 3 and 3x + y – 7 = 0 and which passes through the point (1, 1).
 

Solution:

 The equation of the hyperbola differs from the equation of the asymptotes by a constant.
 
⇒ The equation of the hyperbola with asymptotes 3x + y – 7 = 0 and 2x – y = 3 is (3x + y – 7) (2x – y – 3) + k = 0. It passes through 
(1, 1) ⇒ k = –6.
 
Hence the equation of the hyperbola is (2x – y – 3)(3x + y – 7) = 6.
 
Illustration:
 
Find the angle between the asymptotes of the hyperbola x2/a2–y2/b2 = 1, then length of whose latus rectum is 4/3 and which passes through the point (4, 2).
 
Solution:
 
We have 2b2/a = length of the latus rectum = 4/3 ⇒ 3b2 = 2a
 
Also, the hyperbola passes through the point (4, 2).
 
Hence 16/a2 – 4/b2 = 1 ⇒ 16/a2 – 6/a = 1
 
Or a2 + 6a – 16 = 0 ⇒ (a – 2)(a + 8) = 0 ⇒ a = 2 ⇒ b2 = 4/3.
 
The asymptotes of the given hyperbola are y = + b/a x or y + 1/√3 x.
 
If θ1 and θ2 are the angles which the asymptotes make with the positive x-axis, then
 
        tan θ1 =  ⇒ θ1 = π/6 and tan θ2 = –1/√3 ⇒ θ2 = –π/6.
 
Hence the angle between the asymptotes = π/3.
 
Illustration:
 
Prove that the chords of the hyperbola x2/a2–y2/b2 = 1, which touch its conjugate hyperbola are bisected at the point of contact.
 
Solution:
 
Let P(x1, y1) be the mid-point of the chord of the given hyperbola, so that the equation of the chords is xx1/a2–yy1/b2 = x12/a2–y12/b2.                …… (1)
 
If touches the conjugate hyperbola x2/a2–y2/b2 = 1, then
 
x2/a2–1/b2 [xx1/a2–x12/a2+y12/b2]2.b4/y12 + 1 = 0 will have equal roots. Simplifying, we find that
 
x2/a2 [y12/b2–x12/a2]+2xx1/a2[x12/a2–y12/b2]–[y12/b2–x12/a2]2+y12/b2 = 0 has equal roots so that
 
4x12/a2 [x12/b2–y12/a2]+4[y12/b2–x12/a2][[y12/b2–x12/a2]2–y12/b2]= 0
 
or, x12/a2 [x12/a2–y12/b2]–[x12/a2–y12/b2]2 + y12/b2 = 0 or (x12/a2–y12/b2)(x12/a2–x12/a2+y12/b2)+y12/b2= 0
 
or x12/a2 – y12/b2 + 1= 0 ⇒ (x1, y1) lies on the conjugate hyperbola.
 
Hence the chord (1) touches the conjugate hyperbola at its midpoint (x1, y1).

 
Alternative Solution

 Any tangent to the conjugate hyperbola x2/a2–y2/b2 = –1 is

x = my + √b2m2–a2.                                           …… (2)

If this is same as the chord (1), then m = a2y1/b2x1 and hence

a4/x12 [x12/a2 – y12/b2]2 = b2m2 – a2 = b2a4y12/b4x12 – a2

Or [x12/a2 – y12/b2]2 = y12/b2 – x12/a2 or x12/a2 – y12/b2 = –1

⇒ (x1, y1) lies on the conjugate hyperbola.

⇒ the chord (1) touches conjugate hyperbola and is bisected at the point of contact.
 

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Course Features

  • 53 Video Lectures
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  • Mind Map
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Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution