If any point on an ellipse is (3 α, 5 α) , then what is the eccentricity of the ellipse?

Correct answer: B. (4)/(5) Let x = 3 α and y = 5 α . (x)/(3) = α and (y)/(5) = α Squaring and adding both equations: (x^2)/(9) + (y^2)/(25) = ^2α + ^2α = 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 = (9)/(25) e^2 = (16)/(25) e = (4)/(5) Answer: (4)/(5)

NDA 2025

Exam: NDA
Year: 2025
Subject: Mathematics
Chapter: Ellipse

NDA 2025 · Mathematics · Ellipse

If any point on an ellipse is (3 α, 5 α) , then what is the...

If any point on an ellipse is

(3\sin\alpha, 5\cos\alpha)

, then what is the eccentricity of the ellipse?

Options

A $\dfrac{4}{3}$
B $\dfrac{4}{5}$
C $\dfrac{3}{4}$
D $\dfrac{1}{2}$

Correct answer

B. $\dfrac{4}{5}$

Step-by-step 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}

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