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Question

Eccentricity of ellipse $$\frac{{{x^2}}}{{{a^2} + 1}} + \frac{{{y^2}}}{{{a^2} + 2}} = 1\,is\,\frac{1}{{\sqrt 3 }}$$ then length of Latus rectum is 


A
23
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B
43
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C
23
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D
32
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Solution

The correct option is B $$\frac{4}{{\sqrt 3 }}$$
Let $${ A }^{ 2 }={ a }^{ 2 }+1$$,  $${ B }^{ 2 }={ a }^{ 2 }+2$$    {Here, $${ B }^{ 2 }>{ A }^{ 2 }$$}
So, $$e=\sqrt { 1-\dfrac { { A }^{ 2 } }{ { B }^{ 2 } }  } =\sqrt { 1-\dfrac { { a }^{ 2 }+1 }{ { a }^{ 2 }+2 }  } =\dfrac { 1 }{ \sqrt { { a }^{ 2 }+2 }  } =\dfrac { 1 }{ \sqrt { 3 }  } $$
So,  $$\sqrt { { a }^{ 2 }+2 } =\sqrt { 3 } \Rightarrow a=\pm 1$$
So, length of lotus return $$=2\dfrac { { A }^{ 2 } }{ B } $$
Length $$=\dfrac { 2\left( { a }^{ 2 }+1 \right)  }{ \sqrt { { a }^{ 2 }+2 }  } =\dfrac { 2\left( 2 \right)  }{ \sqrt { 3 }  } =\dfrac { 4 }{ \sqrt { 3 }  } $$

Mathematics

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