1. ## 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!

2. 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?

3. 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?

4. 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.

5. Thank you again Soroban!