$The ends of latus rectum of parabola x^{2} + 8 y = 0 are$

$(- 4, - 2) and (4, 2)$

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$(- 4, - 2) and (4, - 2)$

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$(4, 2) and (- 4, 2)$

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$(4, - 2) and (- 4, 2)$

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Solution

The correct option is B
$(- 4, - 2) and (4, - 2)$

Given,
$x^{2} + 8 y = 0$

$\Rightarrow x^{2} = - 8 y$
On comparing with $x^{2} = - 4 a y$ , we get
$\Rightarrow a = 2$
$So , focus of x^{2} = - 4 a y is (0, - a) Hence focus = (0, - 2)$

The end points of latus rectum of the parabola
$x^{2} = - 4 a y are (- 2 a, - a), (2 a, - a) .$

$Ends of latus rectum = (4, - 2), (- 4, - 2)$ .
Hence, the correct answer is Option c.

Trick: Since the ends of latus rectum lie $on parabola so only points$
$(- 4, - 2) and (4, - 2) satsify the$ parabola.

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