Question

The length of latus rectum of the ellipse 3x² + y² -12x + 2y + 1 = 0 is
1. 2√3
2. 12
3. 4/√3 
4. 3/√2
5. None of these

Answer 1

Correct Answer - Option 3 : 4/√3 

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

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

= 4/√3 units

Hence, option (3) is correct.