The eccentricity of the curve represented by x=3( t+ t), y=4( t- t) is

Correct Answer is : 74{{7}}{4}47

Correct Answer is : 74{{7}}{4}47

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MHTCET 25 Apr 2025 Shift 1 Paper

Question : 103 of 150

Marks: +1, -0

x=3(\cos t+\sin t), y=4(\cos t-\sin t)

Solution:

Concept:

The given parametric equations represent an ellipse after eliminating the parameter

t

Explanation:

Given

x=3(\cos t+\sin t)

and

y=4(\cos t-\sin t)

Rewrite as

\frac{x}{3}=\cos t+\sin t

and

\frac{y}{4}=\cos t-\sin t

Square both equations:

\frac{x^2}{9}=1+\sin 2t

and

\frac{y^2}{16}=1-\sin 2t

Add them:

\frac{x^2}{9}+\frac{y^2}{16}=2

Divide by

2

to get the standard ellipse form:

\frac{x^2}{18}+\frac{y^2}{32}=1

Here,

a^2=18

and

b^2=32

. Since

b^2>a^2

, use

e=\sqrt{1-\frac{a^2}{b^2}}

e=\sqrt{1-\frac{18}{32}}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}

Answer:

The eccentricity is

\frac{\sqrt{7}}{4}

, which is option A.

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