Find the length of latus rectum of $y^{2} + 4 x - 6 y + 13 = 0$

4

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Solution

The correct option is A $4$
$y^{2} + 4 x - 6 y + 13 = 0$
$\Rightarrow (y - 3)^{2} + 4 (x + 1) = 0$
$\Rightarrow (y - 3)^{2} = - 4 (x + 1)$
We know latus rectum of standard parabola $Y^{2} = - 4 a X$ is $4 a$
Comparing given parabola with standard parabola, $X = x + 1, Y = y - 3, a = 1$
Hence latus rectum of the given parabola is $4 a = 4 (1) = 4$

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