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JEE Main 2017

JEE Main 2017 Mathematics question on Conic Sections

The eccentricity of the hyperbola whose length of the latusrectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is
  1. A.
    43\dfrac{4}{3}
  2. B.
    43\dfrac{4}{\sqrt{3}}
  3. C.
    23\dfrac{2}{\sqrt{3}}
  4. D.
    3\sqrt{3}
Show answer and solution

Correct option: C

Solution

For a hyperbola x²/a² - y²/b² = 1, latus rectum = 2b²/a = 8, so b² = 4a.

The conjugate axis = 2b and distance between foci = 2ae. Condition: 2b = (1/2)(2ae), so 2b = ae, meaning b = ae/2.

From b² = a²e² - a² (since b² = a²(e²-1)) and b² = 4a:
4a = a²(e²-1), so 4 = a(e²-1).

From b = ae/2: b² = a²e²/4, so 4a = a²e²/4, giving 16 = ae².

Dividing: ae²/[a(e²-1)] = 16/4 = 4, so e²/(e²-1) = 4, giving e² = 4e² - 4, so 3e² = 4, e = 2/√3.

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