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: $\boxed{\frac{\sqrt{3}}{2} \text{ or } \frac{1}{2}}$