Q111 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
If the length of the major axis of an ellipse is 3 times the length of the minor axis, then its eccentricity is
- A$\dfrac{1}{\sqrt{2}}$
- Correct answerB$\dfrac{2 \sqrt{2}}{3}$
- C$\dfrac{1}{\sqrt{3}}$
- D$\dfrac{2}{\sqrt{3}}$
Solution
Let the length of the major axis be $2a$ and the length of the minor axis be $2b$.
Given that the length of the major axis is 3 times the length of the minor axis, we have $2a = 3(2b)$, which simplifies to $a = 3b$.
The eccentricity $e$ of an ellipse is given by the formula $e = \sqrt{1 - \dfrac{b^2}{a^2}}$.
Substituting $a = 3b$ into the formula:
$e = \sqrt{1 - \dfrac{b^2}{(3b)^2}}$
$e = \sqrt{1 - \dfrac{b^2}{9b^2}}$
$e = \sqrt{1 - \dfrac{1}{9}}$
$e = \sqrt{\dfrac{8}{9}}$
$e = \dfrac{\sqrt{8}}{3} = \dfrac{2\sqrt{2}}{3}$
Answer: $\dfrac{2 \sqrt{2}}{3}$
Given that the length of the major axis is 3 times the length of the minor axis, we have $2a = 3(2b)$, which simplifies to $a = 3b$.
The eccentricity $e$ of an ellipse is given by the formula $e = \sqrt{1 - \dfrac{b^2}{a^2}}$.
Substituting $a = 3b$ into the formula:
$e = \sqrt{1 - \dfrac{b^2}{(3b)^2}}$
$e = \sqrt{1 - \dfrac{b^2}{9b^2}}$
$e = \sqrt{1 - \dfrac{1}{9}}$
$e = \sqrt{\dfrac{8}{9}}$
$e = \dfrac{\sqrt{8}}{3} = \dfrac{2\sqrt{2}}{3}$
Answer: $\dfrac{2 \sqrt{2}}{3}$
Hence, the correct answer is option: 'B'.