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📐 Mathematics  ·  Conic Sections  ·  JEE

Find the eccentricity of the hyperbola whose latus rectum is equal to half of its transverse axis.

Answer: e = sqrt(6)/2.

  • A e = 3/2
  • B e = sqrt(6)/2
  • C e = sqrt(3)/2
  • D e = 5/4

Correct answer: B. e = sqrt(6)/2

Explanation: Latus rectum = 2b<sup>2</sup>/a, transverse axis = 2a. Given 2b<sup>2</sup>/a = (1/2)(2a) = a, so 2b<sup>2</sup> = a<sup>2</sup>, meaning b<sup>2</sup> = a<sup>2</sup>/2. Then e<sup>2</sup> = 1 + b<sup>2</sup>/a<sup>2</sup> = 1 + 1/2 = 3/2, so e = sqrt(3/2) = sqrt(6)/2.

The Four Conics by EccentricityCircle (e=0)Ellipse (0<e<1)Parabola (e=1)Hyperbola (e>1)

All four conics form a single family distinguished only by eccentricity: a circle is the most "closed" (e=0), an ellipse is an elongated closed curve, a parabola is the borderline open curve (e=1), and a hyperbola has two separate open branches (e>1).

Concept context

Study circles, parabolas, ellipses, and hyperbolas as curves formed by intersecting a plane with a double cone, with their standard equations and key properties.

Read the full Conic Sections notes →