<s>[INST] Is $\frac{x^2}{a^2}+\frac{y^2}{b^2} = \frac{y}{b}$ the equation of an ellipse? Shouldn't it be $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$? 
While solving a question today I came across a locus of the form $$\frac{x^2}{a^2}+\frac{y^2}{b^2} = \frac{y}{b}$$ It was told in my book that it is the equation of an ellipse. How is that so?

Forgive me if I ask this, this may seem really silly (I'm still in high school) but isn't the general equation of an ellipse of this form? 
$$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$$ 
Or is it of the form 
$$ax^2+by^2+2hxy+2gx+2fy+c=0$$ Can someone please confirm this for me.
 [/INST] You can manipulate your equation as so:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac y b = 0$$
Then we see by completing the square on top of the $y$ terms as below,
$$\begin{align}
\frac{y^2}{b^2} - \frac y b &= \frac{y^2 - by}{b^2} \\
&= \frac{y^2 - by + b^2/4 - b^2/4}{b^2} \\
&= \frac{(y-b/2)^2 - b^2/4}{b^2} \\
&= \frac{(y-b/2)^2}{b^2} - \frac 1 4
\end{align}$$
Therefore, the first euqation and your equation is equivalent to
$$\frac{x^2}{a^2} + \frac{(y-b/2)^2}{b^2} = \frac 1 4$$
We can multiply throughout by $4$, and in the fractions this is equivalent to multiplying the denominator by $1/4$. Then we have
$$\frac{x^2}{(a/2)^2} + \frac{(y-b/2)^2}{(b/2)^2} = 1 \tag{1 }$$
A slight amending of your definition of an ellipse: an ellipse, centered at $(h,k)$, with semi-major and semi-minor axes $a,b$ which are parallel to the $x$ and $y$ axes, has the form
$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$
In this case, then, your equation - and thus $(1)$ - describes an ellipse, with center $(0,b/2)$, and semi-major and semi-minor axes $a/2,b/2$.
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