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The equation $\frac{x^2}{2-a}+\frac{y^2}{a-5} +1 = 0$ represents an ellipse if $a\; \epsilon$
(A) $(2,\frac{3}{2})\;\cup\;(\frac{3}{2},5)$
(B) $(2,\frac{3}{2})$
(C) $(1,\frac{3}{2})$
(D) $(\frac{3}{2},5)$

This is what I have done,

For representing an ellipse, $$e<1$$ $$\implies \sqrt{1-\frac{a-5}{2-a}} <1$$ $$\implies 1-\frac{a-5}{2-a}<1$$ $$\implies \frac{a-5}{2-a}>0$$ $$\implies \frac{a-5}{a-2}<0$$

So, $a\;\epsilon\;(2,5)$
But there are no such options. So, have I missed a solution ? I think there is a mistake in the way I have solved the problem. Can anybody tell me what to do?

Ris97
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3 Answers3

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you should exclude the case of $e=0$, which is a circle but not a ellipse. So the answer is A.

Additional notes

I think the $3/2$ in all your options should be $7/2$. Otherwise $(2,3/2)$ does not make any sense.

MoonKnight
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  • Why did you choose $\frac 72$? Because it is the center of the interval $(2,5)$? Note that if all the $\frac32$ points are misprints and a valid value is inserted equally into each, then $(C)$ is the only remaining incorrect answer... – abiessu Dec 14 '13 at 06:25
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    @abiessu (1)Because $7/2$ makes it a circle. (2)I agree with you the problem asks only for sufficient conditions, so in that sense B,D are also correct answer. (3) In general actually I think the definition of ellipse include the case of a circle. (4) My understanding is that this is just a poorly phrased problem.... – MoonKnight Dec 14 '13 at 06:31
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    You are right, it should be $\frac{7}{2}$. And it's not my fault that it's wrong! – Ris97 Dec 14 '13 at 16:34
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If

$$\frac {x^2}{2-a}+\frac{y^2}{a-5}+1=0$$

is claimed to be an ellipse, then in standard form,

$$\frac {x^2}{a-2}+\frac{y^2}{5-a}=1$$

which matches exactly with the result that you found, namely that $a\in(2,5)$.

The listed answer in $(A)$ is not correct as written, but the writing of it suggests that perhaps it is written incorrectly and might be something like $(2,\frac 52)\cup (\frac 52,5)$. But if that were the case, then (presumably) $(B),(D)$ would also have the same mixup and would also be correct answers.

Basically, you found the correct interval for $a$, I would go back to the source of the question and make sure that nothing is misprinted or mis-copied...

abiessu
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  • Yes, I thought so too, because $\frac{3}{2}$ is not giving me equal denominators but $\frac{7}{2}$ is giving me equal denominators which, in turn gives me a circle. As, we are considering a general ellipse, so, we have to exclude $\frac{7}{2}$ from the solution set. I think what MoonKnight has written may be right. – Ris97 Dec 14 '13 at 16:31
  • Also, note that $\frac 72=3\frac 12$ which may have contributed to the confusion initially... – abiessu Dec 14 '13 at 16:36
  • That just means that someone (along the chain of you receiving the problem to solve) may have looked at $3\frac12$ on one sheet and incorrectly written $\frac32$ on another... – abiessu Dec 14 '13 at 16:39
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If we agree in $(2,3/2)=\emptyset$ and notice that there is an if and no iff, then there are two interpretations depending on we consider a circle as an ellipse or not. In the first case A, C, and D are correct, otherwise C is the only correct solution.

Michael Hoppe
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  • Umm, I think $(C)$ is the only interval written correctly, and it is non-empty and disjoint with the solution space for $a$ and therefore cannot be the correct answer... – abiessu Dec 14 '13 at 15:33