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A conic with equation $$ a x^2 + b y^2 = c $$ has two focus points, where $a=4$, $b=24$ and $c=65$. One of those focus points has a positive x-coordinate. To two decimal places, what is the value of that positive x-coordinate?

I got the answer as 14.83, seems a bit too big is that right? The above answer is according to this in my textbook enter image description here

enter image description here

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    title:"Just convert this to radians?" the question seems to be asking for a coordinate, not an angle...2012-10-03
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    @Navin Tks for pointing that out my old question was in my browsers cache, changed the title2012-10-03

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Put the ellipse in standard form. Start from $4x^2+24y^2=65$. Divide through by $65$. We get $\frac{4}{65}x^2+\frac{24}{65}y^2=1$.

This is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a^2=\frac{65}{4}$ and $b^2=\frac{65}{24}$. Finally, you want $\sqrt{a^2-b^2}$.

Your intuition is right: this is substantially smaller than your answer. To two decimal places, I get $3.68$.

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    This is an elipse in standard position so the eccentricity is squareroot(1- (a^2/b^2). Correct?2012-10-03
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    Also take note the question is asking for focal point2012-10-03
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    I found the $x$-coordinate of the focal point with positive $x$-coordinate, as the question asks. The eccentricity is not $a^2/b^2$, it is $\frac{\sqrt{a^2-b^2}}{a}$. If your book/notes do not have the relevant formulas, I am sure you can find them in the wikipedia article on the ellipse, there must be one!2012-10-03
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    andre please see my textbook picture, in the original post. If I am wrong please let me know. You and I agree to certain extent but the eccentricity I am not sure that is the formula that you want me to use2012-10-03
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    That is equivalent to what I wrote. But I guess if you follow the book, you will first find $e$, which is $\sqrt{1-\frac{b^2}{a^2}}$ (mine was different but equivalent) and then you will multiply by $a$. You will get $\sqrt{a^2-b^2}$. Anyway, same answer. Eccentricity is about $0.91287$.2012-10-03
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    I agree with you on the eccentricity. But my 14.83 for focal point is still wrong?2012-10-03
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    Yes it is. Remember (if you want to use formula in book, which is more trouble than what I used) that $a=\sqrt{65/4}$. I just figured out what you did wrong. You used $a=65/4$, and it is $a^2$ which is $65/4$. So $a$ is about $4.031$. By the way, you really should change the title, it will mislead people using the search engine.2012-10-03
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    ah yes now i understand updated my question with full working, u always got some tricks up your sleeve. :)2012-10-03