Question

Let the distance between the foci of an ellipse \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a>b) \] be \(4\) and the distance between its directrices be \(10\). Then the length of its latus rectum is:

• A. \( \dfrac{12}{\sqrt{10}} \)
• B. \( \dfrac{6}{\sqrt{10}} \)
• C. \( \dfrac{8}{\sqrt{5}} \)
• D. \( \sqrt{10} \)

Hint:

For ellipse problems, memorize standard results for foci, directrices, eccentricity, and latus rectum to quickly link given data.

Answer 1

The Correct Option is A

To find the length of the latus rectum of the given ellipse, proceed step by step. 

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Step 1: Standard form of ellipse

The standard equation of an ellipse is:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a>b) \]

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Step 2: Use given focal distance

The distance between the foci of an ellipse is \(2c\), where:

\[ c = \sqrt{a^2 - b^2} \]

Given:

\[ 2c = 4 \Rightarrow c = 2 \]

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Step 3: Use distance between directrices

The distance between the directrices of an ellipse is:

\[ \frac{2a^2}{c} \]

Given this distance is 10:

\[ \frac{2a^2}{2} = 10 \]

Solving:

\[ a^2 = 10 \]

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Step 4: Find \(b^2\)

Using:

\[ c = \sqrt{a^2 - b^2} \]

Substitute known values:

\[ 2 = \sqrt{10 - b^2} \]

Squaring both sides:

\[ 4 = 10 - b^2 \Rightarrow b^2 = 6 \]

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Step 5: Length of latus rectum

The length of the latus rectum of an ellipse is given by:

\[ \text{Latus rectum} = \frac{2b^2}{a} \]

Substitute values:

\[ a = \sqrt{10}, \quad b^2 = 6 \]

\[ \text{Length} = \frac{2 \times 6}{\sqrt{10}} = \frac{12}{\sqrt{10}} \]

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Final Answer

\[ \boxed{\frac{12}{\sqrt{10}}} \]