Intuitively, what exactly does the ellipse equation mean?

Question

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?

Answer 1

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.

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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.

Answer 2

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

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

Answer 3

The circle equation is simply one specific ellipse equation, where the scaling of each direction ($x$ and $y$) is identical.

Division by some factor of just one variable matches scaling by that factor in that direction. Hence, $\frac{x^2}{3^2}+y^2=r^2$ will make the ellipse twice as large in the $x$ direction as its $y$ direction - half of $x$ is the same size as a whole $y$.

From here you have variation on the exact size of the scaling for a given radius, and since a scaled scalar is still a scalar, you only require a scalar for each direction, with the original radius set to 1, to obtain the general ellipse equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\space(= 1^2)$$

Note that because the scaling applied to the variable, it ends up also squared when that variable appears in the equation.

If you want to imagine it as the triangular form arising from rotation of the fixed hypotenuse length while variables record the length of each other side, you can consider inclining the plane that the rotation occurs in, and taking the projection of the sides onto the original plane.

Answer 4

Key thing to keep in mind is that the length of the rope doesn't change. This means you're basically modeling all the points whose "sum of distances from the given two fixed points" doesn't change.

Say $A$ and $B$ are the fixed points and $L$ is the length of the rope, then the point P traces the curve given by the equation : $$\text{(distance between P and A)} + \text{(distance between P and B)}= L$$

Try plugging in $P = (x,y)$, $A = (-c,0)$ and $B=(c,0)$ and see what you get.

Answer 5

$x^2+y^2 = 1$ is our unit circle

$\frac {x}{a}$ dilates $x$ by a factor of $a.$ That is it stretches everything horizontally by a factor of $a$

Simmilarly, $\frac {y}{b}$ dilates $y$ by a factor of $b.$

$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\\ \left(\frac xa\right)^2 + \left(\frac yb\right)^2 = 1$

The equation of an ellipse just shows how to distort a circle.

While we can derive the distance between the foci, and the "length of the rope" it is not entirely obvious from the equation.

The length of the major and minor axis $(2a, 2b)$ are.

Answer 6

(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.

Answer 7

The equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ tells one which points belong to the ellipse and which don’t. Take a point with coordinates $(x, y)$, calculate the above. Is it 1? Excellent, it’s on the ellipse. Not 1? Then it’s some other point.

If you want to transform the equation into something that actually provides a method of drawing the shape, try converting it to parametric form. That is, instead of $F(x, y)=0$, look for ways of expressing the same as

$$ \left\{\begin{align*} x &= x(t) \\ y &= y(t). \end{align*}\right. $$

For the ellipse, the formula tells that squares of something add up to 1. What are common things that have this property? Sine and cosine. So $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ is equivalent to $$ \left\{\begin{align*} x &= a \cos t \\ y &= b \sin t \end{align*}\right. $$ in the sense that for any $x$ and $y$, some $t$ exists satisfying the above if and only if $(x, y)$ lies on the ellipse.

But now you actually have a method of drawing the ellipse. Set $t = 0$ and determine $x$ and $y$. Then increase $t$ in small steps and you’ll get more and more points of the ellipse (and when $t$ reaches $2\pi$, you’re done). You can think of $t$ as the time parameter and the equations for $x$ and $y$ as describing the motion of some mechanism that’s drawing the shape. This particular mechanism requires $2\pi$ worth of time to draw the entire ellipse, if you’re wondering how fast and in what direction it’s going at any particular point you can calculate some derivatives and so on.

As for the method of drawing with a string pinned at the foci, I’m not aware of an intuitive reasoning that would explain why this results in the same shape as one gets stretching a circle in one direction. It can be proven mathematically though.

Answer 8

Draw two concentric circles of centre $O$ and radii $a$ and $b$ ($a<b$) in an orthonormal system of axes $(O,\textbf{i},\textbf{j})$. Every half-line from $O$ cuts the circles at points $M_a$ and $M_b$. Let $N_a$ and $N_b$ be the orthogonal projections of $M_a$ and $M_b$ over the $x$ axis and the $y$ axis respectively. Denote $\alpha$ and $\beta$ the measures of the angles $\widehat{N_aIM_a}$ and $\widehat{N_bIM_b}$.

Since $\alpha+\beta=\dfrac{\pi}{2}$ then $$\cos^2\alpha+\cos^2\beta=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$

The ellipse is the set of points $M$ intersection of the lines $(M_aN_a)$ and $(M_bN_b)$.

Answer 9

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.)

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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.

Answer 10

I think it is foci and the semi axes confusion. Intuitively you understand now equation with a and b mentioned above. That is ellipse. Now, because of symmetry, there MUST be a distance from center i. e. foci distance, where the "constant rope length" condition is satisfied.

In reverse you can think from fixed length and two points must drow some shape and you derive ellipse.

However foci distance and a and b are not in trivial intuitive relation.

So by thinking from circle with one parameter, you have now two (a, b semi axes)+one in foci distance to satisfy the fixed rope condition.

Both "intuitive" conditions (fixed rope, foci AND a, b semi axes) collapse into one when going from ellipse to circle. Because the length of the rope is equal to radious. Voila