The directrix of the parabola $x^{2} - 4 x - 8 y + 12 = 0$ is:

y = 0

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x = 1

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y = - 1

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x = - 1

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Solution

The correct option is C $y = - 1$
Given equation of parabola is
$x^{2} - 4 x - 8 y + 12 = 0$
$\Rightarrow x^{2} - 4 x + 4 = 8 y - 12 + 4$
$\Rightarrow (x - 2)^{2} = 8 (y - 1)$
It is a parabola opening upward and vertex is at $(2, 1)$
Latus rectum $4 a = 8$
$\Rightarrow a = 2$
Also, focus is at $(2, 3)$
So, equation of directrix is parallel to x-axis i.e. $y = - 1$ (Distance of vertex from directrix is same as that from focus)

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Similar questions

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

y^{2} = 4 x

x^{2} = 4 y

Find the equation of the parabola whose:

(i) focus is (3,0) and the directrix is 3x+4y=1

(ii) focus is (1,1) and the directrix is x+y+1=0

(iii) focus is (0,0) and the directrix is 2x-y-1=0

(iv) focus is (2,3) and the directrix is x-4y+3=0

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