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Thread: Help with Ellipses

  1. #1
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    Help with Ellipses

    Hello, I have the basics of solving for foci, vertices, eccentricity and graphing ellipses down, but for some reason these two problems have me scratching my head.
    x^2+4y^2=1 and y^2=1-2x^2 both of these need to be put into std form which is (x^2\a^2)+(y^2\b^2)=1, before I can solve for anything. Help!
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  2. #2
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    Hello, DividingBy0!

    $\displaystyle x^2+4y^2\:=\:1\;\text{ and }\;2x^2+y^2 \:=\:1$

    Both of these need to be put into standard form: .$\displaystyle \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} \:=\:1$

    I'll do the first one . . .

    $\displaystyle \displaystyle x^2 + 4y^2 \:=\:1 \quad\Rightarrow\quad \frac{x^2}{1} + \frac{y^2}{\frac{1}{4}} \:=\:1$

    So we have: .$\displaystyle a = 1,\;b = \frac{1}{2}$

    Got it?

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  3. #3
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    Thank you! The follow up question I have for you is why is the denom. 1/4? The second eq in std form would then be $\displaystyle x^2/(1/2)+y^2/1=1$ with a=1 and b=sqrt 1/2 correct?
    Last edited by DividingBy0; Dec 6th 2010 at 07:55 PM.
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  4. #4
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    Hello, DividingBy0!

    $\displaystyle \text{The follow-up question I have for you is: why is the denominator }\frac{1}{4}\,?$

    We had: .$\displaystyle x^2 + 4y^2 \:=\:1 $

    It is almost in standard form: the right side is 1, $\displaystyle \,x^2$ is over $\displaystyle 1^2.$

    But the "4" should not be next to the $\displaystyle \,y^2$, it should be under.


    One way to eliminate the 4 is to multiply top and bottom by $\displaystyle \frac{1}{4}\!:$

    . . $\displaystyle \dfrac{\frac{1}{4}}{\frac{1}{4}}\cdot\dfrac{4y^2}{ 1} \;=\;\dfrac{y^2}{\frac{1}{4}} $




    $\displaystyle \text{The second eq in std form would then be: }\:\dfrac{x^2}{\frac{1}{2}}+ \dfrac{y^2}{1} \:=\:1$

    $\displaystyle \text{with }a\,=\,1\,\text{ and }\,b\,=\,\sqrt{\frac{1}{2}}\quad\hdots\;\text{ Correct?}$

    If you defined $\displaystyle \,a$ to be the larger of the two numbers, this is correct.

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  5. #5
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    Thank you again Soroban!
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