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Thread: Conic Sections Question

  1. #1
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    Question Conic Sections Question

    Hi. Is there any way to know if a ellipse in the form $\displaystyle ax^2+by^2+cx+dy+e=0$ has a horizontal or vertical major axis without changing into standard form. For example, if I have the ellipse $\displaystyle 2x^2+3y^2-x+6y-2=0$ can I tell if it is horizontal or vertical without changing it into standard form. I would think that the quadratic term ($\displaystyle ax^2, by^2$) with the lower coefficent would be over $\displaystyle a^2$ , allowing one to determine whether it was horizontal or vertical, but I could be wrong. Could someone tell me if I am correct? Thanks!
    Last edited by sleigh; Mar 24th 2010 at 02:35 PM. Reason: typed s instead of x
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  2. #2
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    You're on the right track. The coefficients of $\displaystyle x^2\text{ and }y^2$ (a and b) are all that matter. If a>b, it's one, and if a<b it's the other. Of course, if a=b, it's a circle.

    I'll let you figure out which one is which. Look at the equation $\displaystyle ax^2+by^2=\text{constant}$ when x=0 and when y=0.

    Of course, a and b have to be positive, and it's possible that you'll end up with $\displaystyle ax^2+by^2=-\text{constant}$ after completing the square in x and y, so your equation would have no solutions.

    Post again in this thread if you're still having trouble.
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  3. #3
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    I think that whichever one, $\displaystyle x^2$ or $\displaystyle y^2$ has a smaller coefficent would be over $\displaystyle a^2$. In this ellipse: $\displaystyle 2x^2+4y^2=9$, dividing by 9 yields $\displaystyle \frac{2x^2}{9} + \frac{4y^2}{9}=1$. Moving the coefficents of $\displaystyle x^2$ and $\displaystyle y^2$ into the denominator, I get $\displaystyle \frac{x^2}{\frac{9}{2}}+\frac{y^2}{\frac{9}{4}}=1$ Am I correct in assuming that if the coefficent in front of $\displaystyle x^2$ is smaller than the coefficent in front of $\displaystyle y^2$, then the ellipse is horizontal, and if the coefficent in front of $\displaystyle y^2$ is larger than the coefficent in front of $\displaystyle x^2$, it is vertical? Thank you for your help!
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  4. #4
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    You can work straight from $\displaystyle 2x^2+4y^2=9$. If x=0, y is $\displaystyle \pm\frac{3}{2}$, and if y=0, x is $\displaystyle \pm\frac{3}{\sqrt{2}}$. So the horizontal (y=0) axis is longer.

    So you are correct. If the coefficient of $\displaystyle x^2$ is smaller, the major axis is horizontal, and if the coefficient of $\displaystyle y^2$ is smaller, the major axis is vertical.

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