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

$9 x^{2}-6 x+36 y+19=0$

বা, $9\left(x^{2}-\frac{2}{3} x+\frac{4}{9}\right)=-36 y-19+4$

বা, $9\left(x-\frac{2}{3}\right)^{2}=-36\left(y+\frac{15}{36}\right)$

$\begin{aligned} \therefore\left(x-\frac{2}{3}\right)^{2}= & -4\left(y+\frac{15}{36}\right) \end{aligned}$

Latus rectum $=|4 a|=|4 x-1|=4$