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Eccentricity of (x-3)^2+(y-4)^2 = y^2/9 | JEE

JEE Maths question with a full step-by-step solution.

Question

The eccentricity of the ellipse

(x-3)^2+(y-4)^2=\dfrac{y^2}{9}

is:

\dfrac{\sqrt3}{2}

\dfrac{1}{3}

correct

\dfrac{1}{3\sqrt2}

\dfrac{1}{\sqrt3}

Solution

Step 1: Multiply through by

9

and expand.

9(x-3)^2+9(y-4)^2=y^2 \;\Rightarrow\; 9(x-3)^2+9y^2-72y+144=y^2.

\Rightarrow 9(x-3)^2+8y^2-72y+144=0.

Step 2: Complete the square in

y

. Since

8y^2-72y=8\left(y-\tfrac92\right)^2-162

9(x-3)^2+8\left(y-\tfrac92\right)^2=162-144=18.

Step 3: Divide by

18

to reach standard form.

\frac{(x-3)^2}{2}+\frac{\left(y-\tfrac92\right)^2}{9/4}=1.

Here the larger denominator is

\tfrac94

(under the

y

-term), so

a^2=\tfrac94

and

b^2=2

, with the major axis vertical. Step 4: Apply

e^2=1-\dfrac{b^2}{a^2}

e^2=1-\frac{2}{9/4}=1-\frac{8}{9}=\frac19 \;\Rightarrow\; e=\frac13.

Correct answer: (2)

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