The length of the latus rectum of the ellipse 3x2 + y2 = 12 is:

  1. \(\rm \frac{2}{\sqrt3}\)
  2. \(\rm \frac{4}{3\sqrt3}\)
  3. \(\rm \frac{3}{2\sqrt3}\)
  4. \(\rm \frac{4}{\sqrt3}\)

Answer (Detailed Solution Below)

Option 4 : \(\rm \frac{4}{\sqrt3}\)
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NDA 01/2025: English Subject Test
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Detailed Solution

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Concept:

The length of the latus rectum of the ellipse \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is equal to \(\rm \frac{2a^2}{b}\). (a < b)

Calculation:

Writing the equation of the ellipse in the standard form \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), we get:

\(\rm \frac{x^2}{2^2}+\frac{y^2}{(2\sqrt3)^2}=1\)

∴ a = 2 and b = 2√3.

Here a < b

Length of the latus rectum = \(\rm \frac{2a^2}{b}\) = \(\rm \frac{2(2^2)}{2\sqrt3}\) = \(\rm \frac{4}{\sqrt3}\).

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