Question
Download Solution PDFThe length of latus rectum of the ellipse 3x2 + y2 -12x + 2y + 1 = 0 is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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.Last updated on Jul 8, 2025
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