The polar equation of the conic having latus rectum $12$ and eccentricity $\frac{2}{3}$ is

r (3 + 3 cos θ) = 20

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r (3 + 2 cos θ) = 18

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r (2 + 3 cos θ) = 18

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r (1 + 5 cos θ) = 18

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Solution

The correct option is D $r (3 + 2 cos θ) = 18$
Given latus rectum $= 12 \Rightarrow l = 6$ and $e = \frac{2}{3}$
Thus required conic is,
$\frac{l}{r} = 1 + e cos θ$
$\Rightarrow r (2 + 3 cos θ) = 18$

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\frac{5}{r} = 2 + 3 cos θ + 4 sin θ

For all values of $θ$ , the lines represented by the equation
$(2 c o s θ + 3 s i n θ) x + (3 c o s θ - 5 s i n θ) y - (5 c o s θ - 2 s i n θ) = 0$

\frac{R_{1}}{r_{1}} + \frac{R_{2}}{r_{2}} + \frac{R_{3}}{r_{3}} =

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