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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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