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If $\frac{x^{2}}{⨍(4a)} + \frac{y^{2}}{{⨍(a}^{2} -5)} = 1$ represents an ellipse with major axis as y-axis and $⨍$ is decreasing function then $a \in (\infty, 1)$

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$a \in (- 1, 5)$

$a \in (\infty, 1)$

$a \in (1, 4)$

$a \in (5, \infty)$

answer is D.

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Detailed Solution

${⨍ (a}^{2} -5)>⨍ (4a) a^{2} - 5 < 4 a a^{2} - 4 a - 5 < 0 (a - 5) (a + 1) < 0 \frac{+ - +}{- 1 5} a \in (- 1, 5)$

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