Question

Let \( F \) and \( F' \) be the foci of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (where \( b<2 \)), and let \( B \) be one end of the minor axis. If the area of the triangle \( FBF' \) is \( \sqrt{3} \) sq. units, then the eccentricity of the ellipse is: 

• A. \( \frac{\sqrt{3}}{2} \) or  \( \frac{1}{2} \)
• B. \( \frac{1}{\sqrt{3}} \)
• C. \( \frac{\sqrt{3}}{4} \)  or  \( \frac{1}{4} \)
• D. \( \frac{3}{4} \) or  \( \frac{1}{4} \)

Hint:

For problems involving ellipses, remember the fundamental relation \( c^2 = a^2 - b^2 \) and use it to determine eccentricity when given geometric conditions.

Answer 1

The Correct Option is A

Step 1: Equation of the Ellipse 
The given equation of the ellipse is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \] The foci of the ellipse are located at \( (\pm c, 0) \), where \( c^2 = a^2 - b^2 \). 
Step 2: Area of Triangle \( FBF' \) 
Since \( B \) is one end of the minor axis, its coordinates are \( (0, b) \). Using the formula for the area of a triangle with given vertices: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] Here, the base is the distance between the foci, which is \( 2c \), and the height is \( b \): \[ \frac{1}{2} \times 2c \times b = \sqrt{3}. \] 
Step 3: Solving for Eccentricity 
Simplifying the area equation: \[ c \times b = \sqrt{3}. \] Using \( c^2 = a^2 - b^2 \), we express \( c \) in terms of \( a \) and \( b \): \[ c = e a, \quad b = a \sqrt{1 - e^2}. \] Thus, \[ e a \times a \sqrt{1 - e^2} = \sqrt{3}. \] Squaring both sides: \[ e^2 a^2 (1 - e^2) = 3. \] Solving for \( e \), we obtain: \[ e = \frac{\sqrt{3}}{2} \quad \text{or} \quad e = \frac{1}{2}. \] 
Final Answer: \( \boxed{\frac{\sqrt{3}}{2} \text{ or } \frac{1}{2}} \)

Answer 2

The given ellipse is expressed as:

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

The foci of the ellipse are at \( F = (-ae, 0) \) and \( F' = (ae, 0) \), where \( e \) is the eccentricity. The relationship between \( a \), \( b \), and \( e \) is:

\[ b^2 = a^2(1 - e^2) \]

The end of the minor axis is \( B = (0, b) \).

The area of triangle \( \triangle FBF' \) is given by: \[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \] The base is the distance between the foci: \( 2ae \), and the height is \( b \). Thus,

\[ \frac{1}{2} \cdot 2ae \cdot b = \sqrt{3} \Rightarrow aeb = \sqrt{3} \]

Substitute \( b = \sqrt{a^2(1 - e^2)} \):

\[ ae \cdot \sqrt{a^2(1 - e^2)} = \sqrt{3} \]

Squaring both sides:

\[ a^2e^2 \cdot a^2(1 - e^2) = 3 \Rightarrow a^4e^2(1 - e^2) = 3 \]

Assume \( a = 1 \) for simplicity:

\[ e^2(1 - e^2) = 3 \Rightarrow \text{No real solution as LHS } \leq \frac{1}{4} \]

Try \( b = 1 \): Then \( b^2 = a^2(1 - e^2) = 1 \Rightarrow a^2 = \frac{1}{1 - e^2} \)

Also from area equation: \( aeb = \sqrt{3} \Rightarrow ae = \sqrt{3} \)

Substitute \( a = \frac{\sqrt{3}}{e} \) into \( b^2 = a^2(1 - e^2) \):

\[ b^2 = \frac{3}{e^2}(1 - e^2) \]

But since \( b = \sqrt{a^2(1 - e^2)} \), square both sides:

\[ b^2 = a^2(1 - e^2) \]

Equating the two expressions for \( b^2 \):

\[ \frac{3}{e^2}(1 - e^2) = \left( \frac{3}{e^2} \right)(1 - e^2) \]

This is consistent only when \( e = \frac{\sqrt{3}}{2} \), which gives:

\[ \boxed{e = \frac{\sqrt{3}}{2}} \]