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?

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    $\begingroup$ Which ellipse equation are you asking about? There are quite a few to choose from. $\endgroup$
    – amd
    Commented Jun 23, 2018 at 5:27
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    $\begingroup$ a circle equation can be understand very intuitively Then think of it the other way around. An ellipse is what a circle looks like if "dilated" along the axes. The canonical circle equation $\,x^2+y^2=r^2\,$ then becomes something like $\,\lambda x^2+\mu y^2=d^2\,$ i.e. the canonical ellipse equation. $\endgroup$
    – dxiv
    Commented Jun 23, 2018 at 5:28
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    $\begingroup$ @dxiv I see. But then how does the ellipse equation model the process used to create an ellipse, namely, by stretching a string fixed to 2 points and moving it around in a circle (or should I say ellipse). I don't get how the ellipse equation models this. $\endgroup$
    – Ethan Chan
    Commented Jun 23, 2018 at 5:31
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    $\begingroup$ @EthanChan For that you'll have to get down to some actual calculations. My comment was only about the "intuition" of it. See also Why is the equation of an ellipse (x/a)^2 + (y/b)^2 = 1? $\endgroup$
    – dxiv
    Commented Jun 23, 2018 at 5:36
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    $\begingroup$ Fix a rope of lenght $L$ to the points $A$ and $B$ (the foci : $L > d(A,b)$). Then put a pencil in the rope e draw the curve constrained by the rope in tension. $\endgroup$ Commented Jun 23, 2018 at 12:29

10 Answers 10


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.

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    $\begingroup$ It may be worth mentioning that the "standard" ellipse equation has the major and minor axes parallel to the coordinate axes, while the other three presented forms, if I'm not mistaken, are more general. $\endgroup$
    – zahbaz
    Commented Jun 23, 2018 at 20:43

(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:


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.

  • $\begingroup$ Okay. That's fair. So basically, the ellipse equation just gives a set of points that fufills the requirements of an ellipse, and doesn't actually "draw out the ellipse"? $\endgroup$
    – Ethan Chan
    Commented Jun 23, 2018 at 7:06
  • $\begingroup$ @Ethan Yes, though you could argue that it does draw the ellipse once the set of points it represents is "painted" in the plane. $\endgroup$
    – orion2112
    Commented Jun 23, 2018 at 7:11
  • $\begingroup$ Okay. That's fair. One last question though, if that's the case, then why are we using the geometric properties of an ellipse to derive it's equation? How do we derive the equation (the equation that represents a shape) if the equation will have no relation to the geometric properties of the shape? $\endgroup$
    – Ethan Chan
    Commented Jun 23, 2018 at 7:16
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    $\begingroup$ @Ethan You start with the geometric properties and derive an equation (because that's the natural and easy thing to do). Then once you have the equation, you say that its locus (the set of points) IS the ellipse. Then you can almost "forget" about the underlying geometry (but it most definitely is still there; there is still a link between the algebra and the geometry). Thinking in terms of the locus is just more helpful (sometimes) when you deal with equations or relations for which you don't have an intuition. Hope this clarifies... $\endgroup$
    – orion2112
    Commented Jun 23, 2018 at 7:32

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

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

  • $\begingroup$ I see, but how does this fit in with the process used to derive an ellipse? This makes sense, but why is the process of deriving the ellipse so complicated? Where does this fit in with all of that? $\endgroup$
    – Ethan Chan
    Commented Jun 23, 2018 at 5:29
  • $\begingroup$ Circle is a special case of ellipse where a=b=R. Ellipse is a deformed circle so we scale the coordinates to get an ellipse from the circle. Deriving the equation of ellipse from its definition is complicated but from scaling a circle it is much easier. $\endgroup$ Commented Jun 23, 2018 at 5:33
  • $\begingroup$ I get the scaling of the circle explanation, and I find it helpful. But I also want an intuitive understanding of the equation from how it is derived from the definition. I get how to derive it from the definition, but I don't find the link intuitive. I don't see how the equation shows how an ellipse can be drawn from it's definition, if you see what I mean. I just don't find the link between the definition of ellipse and the equation of ellipse intuitive. I'd be happy to do some calculation to figure it out; I'm just really confused and would benefit from some guidance. $\endgroup$
    – Ethan Chan
    Commented Jun 23, 2018 at 5:57
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    $\begingroup$ The equation does not lend itself to "intuitive" understanding. You just have to get used to it and graph many ellipses with different parameters to develop a good feeling about it. $\endgroup$ Commented Jun 23, 2018 at 6:27
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    $\begingroup$ @Ethan I think this is actually really intuitive, just not the same as the common definition of ellipse. This is an instruction - draw a circle, then stretch the paper and you get an ellipse. $\endgroup$
    – Džuris
    Commented Jun 23, 2018 at 10:30

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.


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.


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 $(M_bN_b)$.

  • $\begingroup$ [+1] Very neat ! $\endgroup$
    – Jean Marie
    Commented Jan 25, 2019 at 11:43

$x^2+y^2 = 1$ 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.


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.

  • $\begingroup$ I'm glad someone made this contribution, as it directly answers what the OP was asking for. $\endgroup$
    – Eli Rose
    Commented Jun 23, 2018 at 22:25

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.


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


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