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The point of intersection of the latus rectum and axis of the parabola $y^{2} + 4 x + 2 y - 8 = 0$ is

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$(\frac{5}{4}, - 1)$

$(\frac{7}{2}, \frac{5}{2})$

$(\frac{- 5}{4}, 1)$

$(\frac{9}{4}, - 1)$

answer is B.

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Detailed Solution

${(y + 1)}^{2} = - 4 (x - \frac{9}{4})$
Focus $S = (- 1 + \frac{9}{4}, - 1)$

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