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Maths-

General

Easy

Question

The conic with length of latus rectum 6 and eccentricity is

$\frac{3}{r} = \sqrt{2} + C o s θ$
$\frac{3}{r} = 1 + \sqrt{2} C o s θ$
$\frac{6}{r} = \sqrt{2} + C o s θ$
$\frac{6}{r} = 1 + \sqrt{2} C o s θ$

The correct answer is: $3 over r equals 1 plus square root of 2 C o s space theta$

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