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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=y29(x-3)^2+(y-4)^2=\dfrac{y^2}{9} is:
A32\dfrac{\sqrt3}{2}
B13\dfrac{1}{3}correct
C132\dfrac{1}{3\sqrt2}
D13\dfrac{1}{\sqrt3}
Solution
Step 1: Multiply through by 99 and expand.
9(x−3)2+9(y−4)2=y2  ⇒  9(x−3)2+9y2−72y+144=y2.9(x-3)^2+9(y-4)^2=y^2 \;\Rightarrow\; 9(x-3)^2+9y^2-72y+144=y^2.
⇒9(x−3)2+8y2−72y+144=0.\Rightarrow 9(x-3)^2+8y^2-72y+144=0.
 Step 2: Complete the square in yy. Since 8y2−72y=8(y−92)2−1628y^2-72y=8\left(y-\tfrac92\right)^2-162,
9(x−3)2+8(y−92)2=162−144=18.9(x-3)^2+8\left(y-\tfrac92\right)^2=162-144=18.
 Step 3: Divide by 1818 to reach standard form.
(x−3)22+(y−92)29/4=1.\frac{(x-3)^2}{2}+\frac{\left(y-\tfrac92\right)^2}{9/4}=1.
 Here the larger denominator is 94\tfrac94 (under the yy-term), so a2=94a^2=\tfrac94 and b2=2b^2=2, with the major axis vertical. Step 4: Apply e2=1−b2a2e^2=1-\dfrac{b^2}{a^2}.
e2=1−29/4=1−89=19  ⇒  e=13.e^2=1-\frac{2}{9/4}=1-\frac{8}{9}=\frac19 \;\Rightarrow\; e=\frac13.
 Correct answer: (2)
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