The one end of the latusrectum of the parabola $y^{2} - 4 x - 2 y - 3 = 0$ is at

(0, - 1)

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(0, 1)

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(0, - 3)

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(3, 0)

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(0, 2)

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Solution

The correct option is A $(0, - 1)$
Given equation is $y^{2} - 4 x - 2 y - 3 = 0$
$\Rightarrow (y^{2} - 2 y) = 4 x + 3$
$\Rightarrow {(y - 1)}^{2} - 1 = 4 x + 3$
$\Rightarrow {(y - 1)}^{2} = 4 (x + 1)$
Shift the origin to $(- 1, 1)$
$\Rightarrow Y^{2} = 4 X$
Here focus is at $(1, 0)$ .
Hence, focus of original parabola becomes $(1 - 1, 0 + 1) = (0, 1)$
Therefore, equation of latusrectum is $x = 0$ .
Thus point of intersection of parabola and latus rectum is
$y^{2} - 2 y - 3 = 0 \Rightarrow y = - 1$ or $3$
So, the required points are $(0, - 1), (0, 3)$ .

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List-I	List-II
A) Vertex of the parabola $y^{2} - x + 4 y + 5 = 0$	p) (1,4)
B) Focus of the parabola $x^{2} - 2 x - 12 y + 13 = 0$	q). (1,-2)
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