The length of latus rectum of the ellipse 3x2 + y2 -12x + 2y + 1 = 0 is

  1. \(2\sqrt 3\)
  2. 12
  3. \(\frac{4}{\sqrt 3}\)
  4. \(\frac{3}{\sqrt 2}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{4}{\sqrt 3}\)
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Detailed Solution

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Concept:

Standard Equation of ellipse: \(\frac{{{\rm{\;}}{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} + \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1\)

Length of latus rectum = 2b2/a, when a > b and 2a2/b, when a < b

Calculation:

3x2 + y2 -12x + 2y + 1 = 0

⇒ 3(x2 - 4x + 4) – 12 + (y2 + 2y + 1) = 0

⇒ 3(x – 2)2 – 12 + (y + 1)2 = 0

⇒ 3(x – 2)2 + (y + 1)2 = 12

\( \Rightarrow \frac{{3{{\left( {{\rm{x}} - 2} \right)}^2}}}{{12}} + \frac{{{{\left( {{\rm{y}} + 1} \right)}^2}}}{{12}} = 1\)                                  (Divide by 12)

\( \Rightarrow \frac{{{{\left( {{\rm{x}} - 2} \right)}^2}}}{4} + \frac{{{{\left( {{\rm{y}} + 1} \right)}^2}}}{{12}} = 1\)

\( \Rightarrow \frac{{{{\left( {{\rm{x}} - 2} \right)}^2}}}{{{2^2}}} + \frac{{{{\left( {{\rm{y}} + 1} \right)}^2}}}{{{{\left( {2\sqrt 3 } \right)}^2}}} = 1\)

∴ a2 = 22 and b2 = (2√3)2

Here a < b

So, length of latus rectum = 2a2/b

\(\frac{{2\left( 4 \right)}}{{2\sqrt 3 }}\)

\(\frac{4}{\sqrt 3}\) units

Hence, option (3) is correct.
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