The length of the latus-rectum of the parabola $x^{2} - 4 x - 8 y + 12 = 0$ is

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Solution

The correct option is C
8

Given : $x^{2} - 4 x - 8 y + 12 = 0$

$\Rightarrow (x - 2)^{2} - 8 y + 8 = 0$

$\Rightarrow (x - 2)^{2} = 8 y - 8 = 8 (y - 1)$

Let X=x-2,Y=y-1

$∴ X^{2} = 8 Y$

$∴ Length of the latus rectum = 4 a = 8 u n i t s$

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