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.

• $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ is the general equation of an ellipse with its centre at $(0,0)$ and its axes aligned with the coordinate axes. Your ellipse has centre somewhere else.... Nov 8 '19 at 4:45
• @LordSharktheUnknown But still that would be of the form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1$ right? Nov 8 '19 at 4:47
• Your ellipse does have that form. Nov 8 '19 at 4:48
• @LordSharktheUnknown Isn't there an isolated y at the end? How would I then factor it out to make whole squares? EDIT: Got the answer below Nov 8 '19 at 4:49

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 $$a/2,b/2$$.