<s>[INST] 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 :)!
 [/INST] 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:
$\boxed{y=\pm\sqrt{b^2-\dfrac{b^2x^2}{a^2}}}$
</s>