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When Does x^2/(10-a)+y^2/(4-a)=1 Form an Ellipse | JEE

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Question
The equation x210−a+y24−a=1\dfrac{x^2}{10-a}+\dfrac{y^2}{4-a}=1 represents an ellipse if:
Aa<4a<4correct
Ba>4a>4
C4<a<104<a<10
Da>10a>10
Solution
For the equation to represent an ellipse, both denominators must be positive. Step 1: Require 10−a>010-a>0, which gives a<10a<10. Step 2: Require 4−a>04-a>0, which gives a<4a<4. Step 3: Both conditions hold simultaneously when a<4a<4. The denominators 10−a10-a and 4−a4-a are never equal, so the curve is a genuine ellipse (never a circle) throughout this range. ∴a∈(−∞,4)\therefore a\in(-\infty,4). Correct answer: (1)
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