Intuitively, what exactly does the ellipse equation mean?

I get how to derive the ellipse equation, but I'm struggling to understand what it means intuitively.

You see, a circle equation can be understood very intuitively. The circle equation models how the radius of the circle can be represented using the Pythagorean theorem. But I don't understand what the ellipse equation means at such a level. Does it model how an ellipse can be drawn out using a stretched rope? What exactly does it model? Can someone please explain?

Can you please explain it as simply as possible, as I'm still a beginner?

There is no single equation for an ellipse, just as there is no single equation for a line. We choose a form to highlight information of interest in the current context.

Consider this sampling of ways to write the equation of a line:

$$\begin{array}{rcccl} \text{slope-intercept} &\qquad& y = m x + b &\qquad& \begin{array}{rl} m:&\text{slope} \\ b:&y\text{-intercept} \end{array} \\[8pt] \text{intercept-intercept} && \dfrac{x}{a} + \dfrac{y}{b} = 1 && \begin{array}{rl} a:& x\text{-intercept} \\ b:& y\text{-intercept} \end{array} \\[8pt] \text{normal} && x \cos\theta + y\sin\theta = d && \begin{array}{rl} \theta:& \text{direction of normal} \\ d :&\text{distance from origin} \end{array}\\[8pt] \text{point-slope} && y-y_1= m (x-x_1) && \begin{array}{rl} (x_1,y_1):&\text{point on line} \\ m:&\text{slope} \end{array} \\[8pt] \text{two-point} && \dfrac{y-y_1}{x-x_1}=\dfrac{y_1-y_2}{x_1-x_2} && \begin{array}{rl} (x_i,y_i):&\text{points on line} \end{array}\\[8pt] \text{standard/general} && A x + B y + C = 0 && \end{array}$$

Each form tells us something about the line's geometry. (The "general" form tells us that the line's geometry is unimportant.) Algebra lets us move from one form to another if and when our priorities change.

Note that, since all the forms represent the same line, they must encode the same geometric information somehow. The encodings aren't always neat and tidy, though. For instance, we can manipulate the general form into slope-intercept ... $$A x + B y + C = 0 \qquad\to\qquad y = - \frac{A}{B} x - \frac{C}{B}$$ ... to see that the line's slope is $-A/B$, and its $y$-intercept is $-C/B$. Converting to intercept-intercept form tells us that the $x$-intercept is $-C/A$. Moreover, we can determine slope from the intercept-intercept form, or normal direction from the two-point form, ... whatever. Having the various forms available gives us flexibility in how we present that information. But I digress ...

Likewise, we have a sampling of equational forms for an ellipse.

$$\begin{array}{rcl} \text{foci and string} & \begin{align} \sqrt{(x-x_1)^2+(y-y_1)^2} \qquad&\\ + \sqrt{(x-x_2)^2+(y-y_2)^2} &= 2 a \end{align} & \begin{array}{rl} (x_i,y_i):&\text{foci} \\ 2a:&\text{string length} \end{array} \\[10pt] \text{standard} & \dfrac{(x-x_0)^2}{a^2} + \dfrac{(y-y_0)^2}{b^2} = 1 & \begin{array}{rl} (x_0,y_0):&\text{center} \\ a:&\text{horizontal radius} \\ b:&\text{vertical radius} \end{array}\\[10pt] \text{focus-directrix} & \begin{array}{c} \sqrt{(x-x_0)^2+(y-y_0)^2} \\ \qquad\qquad\qquad = e\;\dfrac{| a x + b y + c |}{a^2+b^2} \end{array} & \begin{array}{rl} (x_0,y_0):&\text{focus} \\ ax+by+c=0:&\text{directrix} \\ e:&\text{eccentricity} \end{array}\\[10pt] \text{general} & A x^2 + B xy + C y^2 + D x + E y + F = 0 & \end{array}$$

The "foci and string" form is the direct (dare I say, "intuitive"?) translation of the foci-and-string definition of the ellipse: the sum of the distances from two points is a constant. We tend not to see that form except as the point of departure on an algebraic journey to the "standard" form. That's because (1) the giant radical expressions are bulky, and (2) the standard form offers much more glance-able information about the geometry of the ellipse, and it has an all-around nicer algebraic nature.

The upshot is that we have an equation to fit every way of looking at an ellipse, so that everyone's intuition is satisfied. And, again, having multiple forms available gives us flexibility in how we want to encode or present the geometric information we find most important to the task at hand.

As an aside, I'll note that the lesser-used focus-directrix form of the equation is more versatile than the standard form, since it works for every conic section (except the circle). In particular, it can be convenient to remember that a parabola (which has eccentricity $1$) has this equation:

$$(x-x_0)^2+(y-y_0)^2 = ( x\cos\theta + y\sin\theta -d )^2$$ where we've leveraged the normal form of the directrix equation to make things tidier.

(This really should be a comment but I needed more space, apologies).

You seem to think (reading the previous comments) that the ellipse's equation should "instruct" us, in a step-by-step manner, how to draw the ellipse. That doesn't have to be the case.

In fact, let us invent a new relation between $x$ and $y$:

$e^x-y=\sin\left(x\cdot y\right)$

There are points $(x,y)$ in the plane that satisfy the above equation and they align on a curve we might as well call a zwiggle. See the WolframAlpha graph or try Desmos or something similar if you are curious what it looks like.

Is it obvious what shape zwiggles look like? No. Does it have to be? No. So... what is a zwiggle? It's just the set of points that satisfy $e^x-y=\sin(x\cdot y)$.

Now, is it obvious what curve we get with:

$\dfrac{(x-2)^2}{81}+\dfrac{(y+1)^2}{25}=1$

Well, to a trained eye, it might be obvious that it is an ellipse centered at $(2,-1)$ with horizontal axis of length $18$ and vertical axis of length $10$ but otherwise, if you don't recognize the equation of an ellipse, you can just tell yourself that the relation encodes a bunch of points in the plane. That "bunch of points" is in fact called the locus of the equation. And in this case, the locus is so popular we have a name for it (ellipse). It just so happens that the ellipse's equation in relation to its shape is something less intuitive than the circle, but more intuitive than the zwiggle.

I know this doesn't quite answer your question (and as I said, this should be a comment), but others have already posted plenty of useful information about ellipses that should give you better intuition. I just hope this helps you see that sometimes, mathematical relations don't really translate to something "geometrically obvious" and you just need to think of the curve as something more abstract.

You may derive the equation for ellipse from the equation of circle by scaling your $x$ and $y$

$$ x^2 + y^2 = R^2 $$

Let $$x\to x/a,\text {and } y\to y/b$$

You get $$ (x/a)^2 + (y/b)^2 = R^2 $$

$$ (\frac {x}{aR})^2 + (\frac {y}{bR})^2 = 1 $$ $$ \frac {x^2}{A^2} + \frac {y^2}{B^2} = 1$$

I mentioned in my previous answer that the "foci and string" form of the ellipse equation is rarely seen "except as the point of departure on an algebraic journey to the 'standard' form". I want to elaborate a little on that "algebraic journey".

Typically, the journey involves a lot of unenlightening, mechanical symbol-pushing to eliminate the square roots. Specifically, defining $$d_i := \sqrt{(x-x_i)^2+(y-y_i)^2}$$ the argument tends to go something like this: $$\begin{alignat}{2}\quad && d_1 + d_2 &= 2 a \qquad\text{(definition: sum of distances to foci is constant)} \tag{$\star$} \\ \to\quad && (d_1 + d_2)^2 &= (2 a)^2 \\[4pt] \to\quad && d_1^2 + 2 d_1 d_2 + d_2^2 &= 4 a^2 \\[4pt] \to\quad && 2 d_1 d_2 &= 4 a^2 - d_1^2 - d_2^2 \\[4pt] \to\quad && (2d_1d_2)^2 &= ( 4 a^2-d_1^2-d_2)^2 \\[4pt] \to\quad && 0 &= d_1^4+d_2^4+16a^4-2d_1^2d_2^2-8a^2d_1^2-8a^2d_2^2 \tag{$\star\star$} \end{alignat}$$ so that $(\star\star)$ contains only even powers of the $d_i$, hence: no radicals. Mission accomplished! Replacing the $d_i$ (in particular, with $(x_i,y_i) = (\pm c,0)$, and defining $b^2 := a^2-c^2$), equation $(\star\star)$ simplifies (see below) to the origin-centered standard form equation we all know and love.

I believe OP is disappointed that, somewhere along the tedious journey from $(\star)$ to $(\star\star)$, we lose sight of $(\star)$.

However, it's still possible to catch a glimpse of $(\star)$ in $(\star\star)$, because $(\star\star)$ factors:

$$(d_1+d_2-2a)(d_1-d_2-2a)(-d_1+d_2-2a)(d_1+d_2+2a) = 0 \tag{$\star\star\star$}$$

(The reader might see a resemblance to Heron's formula in the above.)

Since $(\star)$ is right there in the first factor, the set of points satisfying $(\star\star\star)$ must include those satisfying $(\star)$, the (well, one) definition of the ellipse.

Note that the last factor of $(\star\star\star)$ contributes no points, since presumably $a > 0$ and $d_i \geq 0$.

Interestingly, the middle factors of $(\star\star\star)$ correspond to the relations $$d_1 - d_2 = 2a \quad\text{or}\quad d_2 - d_1 = 2a \qquad\qquad\text{i.e.,}\quad |d_1-d_2| = 2a$$ which say precisely that the difference of distances to the foci is constant: the (well, one) definition of the hyperbola! (Each factor corresponds to an arm of the ostensible hyperbola.)

Consequently, $(\star\star)$ is simultaneously an ellipse equation and an hyperbola equation! Except, not exactly. The graph of the solution set is only one or the other, as determined by $a$'s relationship to the distance between the foci. To be specific, let's do the simplification hinted at earlier: take $(x_i,y_i) = (\pm c,0)$, so that $(\star\star)$ becomes $$16 \left(\;a^2 (a^2 - c^2) - x^2(a^2-c^2) - a^2 y^2\;\right) = 0 \qquad\to\qquad \frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1$$ We see, then, that when $a > c$ ---so that the sum of the distances to the foci is bigger than the distance between the foci themselves--- the equation is that of an ellipse; in $(\star\star\star)$, the second and third factors cannot be zero. On the other hand, when $a < c$, the equation is that of an hyperbola; the first factor of $(\star\star\star)$ cannot be zero. (Exploring the degeneracies arising from $a=c$ is left as an exercise to the reader.)

Anyway, my point is this: We can get to $(\star\star)$ from $(\star)$ by plodding through sequence of algebraic steps that obscure the geometry; or, we can get to $(\star\star)$ by "rationalizing" $(\star)$ via multiplication by what one might call its "Heronic conjugate", the three factors of which are geometrically meaningful (although one is inherently extraneous). And we get the hyperbola equation for free ... because it's the same equation!

Kinda neat, that.