The length of the latus rectum of the parabola $9 x^{2} - 6 x + 36 y + 19 = 0$

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Solution

The correct option is D
4

$9 x^{2}$ - 6x + 19 = -36y

$\Rightarrow (3 x - 1)^{2} = - 36 y - 18 = - 36 (y + \frac{1}{2}) \Rightarrow 9 {(x - \frac{1}{3})}^{2} = - 36 (y + \frac{1}{2})$

Hence length of latus rectum is 4.

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