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JEE Main 2023
Conic Sections
Hyperbola
Hard

Question

Let Hn:x21+n−y23+n=1,n∈N\mathrm{H}_{\mathrm{n}}: \frac{x^{2}}{1+n}-\frac{y^{2}}{3+n}=1, n \in N. Let k\mathrm{k} be the smallest even value of n\mathrm{n} such that the eccentricity of Hk\mathrm{H}_{\mathrm{k}} is a rational number. If ll is the length of the latus rectum of Hk\mathrm{H}_{\mathrm{k}}, then 21l21 l is equal to ____________.

Answer: 2

Solution

Understanding the Hyperbola and its Eccentricity

The problem asks us to analyze a family of hyperbolas HnH_n and find a specific member HkH_k based on its eccentricity. We then need to calculate a property of this hyperbola, its latus rectum.

The standard form of a hyperbola with its transverse axis along the x-axis is given by: x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 where aa is the length of the semi-transverse axis and bb is the length of the semi-conjugate axis. The eccentricity ee of such a hyperbola is defined by the relation:

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