# Finding $y$ value of canonical ellipse.

I have an ellipse: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ This may be a simple question, but my mind plays tricks on me at the moment;

Which is the most efficient way if I have $x$, $a$ and $b$ and want to find the value of $y$?

Hope someone can help me - thanks in advance :)!

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Given:

$\boxed{\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1}$

We can subtract $\dfrac{x^2}{a^2}$ from both sides:

$\dfrac{x^2}{a^2}-\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1-\dfrac{x^2}{a^2}$

Next multiplying both sides by $b^2$:

$\left(\dfrac{y^2}{b^2}\right)b^2=\left(1-\dfrac{x^2}{a^2}\right)b^2$

Which becomes:

$\dfrac{b^2y^2}{b^2}=b^2-\dfrac{b^2x^2}{a^2}$

The b's on the left hand side cancel leaving us with:

$y^2=b^2-\dfrac{b^2x^2}{a^2}$

We can take the square root of both sides leaving us with our final answer:

$\boxed{y=\pm\sqrt{b^2-\dfrac{b^2x^2}{a^2}}}$

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You can rearrange $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ to $$y^2 = b^2(\frac{x^2}{a^2} - 1)$$

With knowledge of $a,b$ and $x$ you can evaluate $y^2$ and $$y_{1,2} = \pm \sqrt{y^2} = \pm \sqrt{b^2(\frac{x^2}{a^2} - 1)}$$

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