Q209 / 269 · NDA 2025 (Phase 1) · Mathematics
If any point on an ellipse is $(3\sin\alpha, 5\cos\alpha)$, then what is the eccentricity of the ellipse?
- A$\dfrac{4}{3}$
- Correct answerB$\dfrac{4}{5}$
- C$\dfrac{3}{4}$
- D$\dfrac{1}{2}$
Solution
Let $x = 3\sin\alpha$ and $y = 5\cos\alpha$.
$\dfrac{x}{3} = \sin\alpha$ and $\dfrac{y}{5} = \cos\alpha$
Squaring and adding both equations:
$\dfrac{x^2}{9} + \dfrac{y^2}{25} = \sin^2\alpha + \cos^2\alpha = 1$
This represents an ellipse with $a^2 = 9$ and $b^2 = 25$.
Since $b > a$, the eccentricity $e$ is given by $a^2 = b^2(1 - e^2)$.
$9 = 25(1 - e^2)$
$1 - e^2 = \dfrac{9}{25}$
$e^2 = \dfrac{16}{25}$
$e = \dfrac{4}{5}$
Answer: $\dfrac{4}{5}$
$\dfrac{x}{3} = \sin\alpha$ and $\dfrac{y}{5} = \cos\alpha$
Squaring and adding both equations:
$\dfrac{x^2}{9} + \dfrac{y^2}{25} = \sin^2\alpha + \cos^2\alpha = 1$
This represents an ellipse with $a^2 = 9$ and $b^2 = 25$.
Since $b > a$, the eccentricity $e$ is given by $a^2 = b^2(1 - e^2)$.
$9 = 25(1 - e^2)$
$1 - e^2 = \dfrac{9}{25}$
$e^2 = \dfrac{16}{25}$
$e = \dfrac{4}{5}$
Answer: $\dfrac{4}{5}$
Hence, the correct answer is option: 'B'.