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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.
 

To read more, Buy study materials of Hyperbola comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.


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