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JEE Previous Year Question

$Foci of the ellipse 2 x^{2} + 3 y^{2} - 4 x - 12 y + 13 = 0 are$

Topic: JEE PYQ

Answer: A. Full step-by-step solution below.

A $(1 + \frac{1}{6}, 2) and (1 - \frac{1}{6}, 2)$
B $(\frac{1}{6} + 1, 2) and (\frac{1}{6} - 1, 2)$
C $(2, 1 + \frac{1}{6}) and (2, 1 - \frac{1}{6})$
D $(2, \frac{1}{6} + 1) and (2, \frac{1}{6} - 1)$

Correct answer: A

Solution

(a) Given ellipse is 2 x^{2} + 3 y^{2} - 4 x - 12 y + 13 = 0 \Rightarrow \frac{( x - 1 ) ^{2}}{( \frac{1}{2} ) ^{2}} + \frac{( y - 1 ) ^{2}}{( \frac{1}{3} ) ^{2}} = 1 Here, Centre (h, k) = (1, 2) Now, e = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{1/3}{1/2} = \frac{1}{3} \Rightarrow a = \frac{1}{2}, b = \frac{1}{3} \Rightarrow ae = \frac{1}{6} Hence, Foci (h \pm ae, k) \equiv (1 \pm \frac{1}{6}, 2)

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