The equation x210-k+y2k-4=1 represents an ellipse then

10-k>0, k-4>0 k 4if given curve ellipse10-k≠k-4 14≠2k-⇒k≠7 k∈(4, 7)∪(7, 10)

The equation x210-k+y2k-4=1 represents an ellipse then

10-k>0, k-4>0 k 4if given curve ellipse10-k≠k-4 14≠2k-⇒k≠7 k∈(4, 7)∪(7, 10)

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The equation $\frac{x^{2}}{10 - k} + \frac{y^{2}}{k - 4} = 1$ represents an ellipse then

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k > 10

4 < k < 10

k < 4

$k \in (4, 7) \cup (7, 10)$

answer is D.

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Detailed Solution

$10 - k > 0, k - 4 > 0 k < 10, k > 4$

if given curve ellipse

$10 - k \neq k - 4 14 \neq 2 k - \Rightarrow k \neq 7 k \in (4, 7) \cup (7, 10)$

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The equation x210-k+y2k-4=1 represents an ellipse then

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The equation $\frac{x^{2}}{10 - k} + \frac{y^{2}}{k - 4} = 1$ represents an ellipse then