Question

The foci of an ellipse are at (-3, 0) and (3, 0). If the eccentricity of the ellipse is 1/2, then the equation of the ellipse is:

• A. $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$
• B. $\dfrac{x^2}{16} + \dfrac{y^2}{7} = 1$
• C. $\dfrac{x^2}{16} + \dfrac{y^2}{25} = 1$
• D. $\dfrac{x^2}{36} + \dfrac{y^2}{16} = 1$
• E. $\dfrac{x^2}{36} + \dfrac{y^2}{27} = 1$

Hint:

For an ellipse: $c = ae$, $b^2 = a^2(1 - e^2)$. Remember $a$ > $b$ > $0$ when the major axis lies along the $x$-axis.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
Use the relations $c = ae$ and $b^2 = a^2 - c^2$ for an ellipse with foci at $(\pm c,\,0)$ and eccentricity $e$.

Step 2: Detailed Explanation:
The foci are $(\pm 3,\,0)$, so $c = 3$ and $e = \dfrac{1}{2}$.
$c = ae \Rightarrow 3 = a \cdot \dfrac{1}{2} \Rightarrow a = 6$, so $a^2 = 36$.
$b^2 = a^2 - c^2 = 36 - 9 = 27$.
Equation: $\dfrac{x^2}{36} + \dfrac{y^2}{27} = 1$.

Step 3: Final Answer:
The equation of the ellipse is $\dfrac{x^2}{36} + \dfrac{y^2}{27} = 1$.