The parametric equation of an ellipse is $$x=a \cos t\\y=b \sin t$$ It can be viewed as $x$ coordinate from circle with radius $a$, $y$ coordinate from circle with radius $b$.


How to prove that it's an ellipse by definition of ellipse (a curve on a plane that surrounds two focal points such that the sum of the distances to the two focal points is constant for every point on the curve) without using trigonometry and standard equation of ellipse?

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    $\begingroup$ How on earth can one work with the question without trigonometry if the given parametrisation is itself a trig function? It is like asking to rescue someone drowning without disturbing the water. $\endgroup$ Jan 7, 2016 at 13:00
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    $\begingroup$ The standard form of the equation is nothing more than the simplification of the sum-of-distances-to-foci-is-constant definition. $$\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a \qquad\to\qquad\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1$$ If you choose to start at the sum-of-distances property, you'll effectively re-trace a bunch of algebraic steps (clearing square roots, etc) from the derivation of the standard equation. That seems like a lot of wasted effort. $\endgroup$
    – Blue
    Jan 7, 2016 at 13:01
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    $\begingroup$ @NeerajKumar You only need to use trigonometry if you want to find a point on the ellipse for a specific value of $t$ in $\sin t$ and $\cos t$. But you don't need to say what $t$ is; just take any arbitrary point $(x,y)$ on the ellipse and use similar right triangles to find an equation relating $x$ and $y$. This is just geometry, not trig. $\endgroup$
    – David K
    Jan 7, 2016 at 13:37
  • $\begingroup$ It is a didactical disaster that an ellipse is commonly defined in the terms given above. In the first place an ellipse is the image of a circle under parallel projection, or an affine image of a circle. It is a miracle that these ovals have the geometrical property described in your definition. $\endgroup$ Jan 7, 2016 at 16:14

2 Answers 2


Let $OA=a$ and $OB=b$ be the radii of the two circles, and let $C$, $C'$ be the foci of the ellipse, where $OC=OC'=c=\sqrt{a^2-b^2}$. If $H$ is the projection of $A$ on major axis $DE$ and $P$ is the projection of $B$ on $AH$, then you want to show that $PC+PC'=2a$.

Suppose, without loss of generality, that $C$ is the focus nearest to $P$. We have $PC^2=PH^2+CH^2$, but $PH=(b/a)AH$, $HC=|c-OH|$ and $AH^2+OH^2=a^2$, so that: $$ \begin{aligned} PC^2&={b^2\over a^2}AH^2+(c-OH)^2={b^2\over a^2}(a^2-OH^2)+c^2+OH^2-2cOH\\ &=a^2+{c^2\over a^2}OH^2-2cOH=\left(a-{c\over a}OH\right)^2,\\ \end{aligned} $$ and then $PC=a-(c/a)OH$. An analogous computation yields $PC'=a+(c/a)OH$, so that $PC+PC'=2a$, QED.

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  • $\begingroup$ This uses the general idea of the solution I was going to recommend, but you did the mechanics of it much better than I did. I like it. $\endgroup$
    – David K
    Jan 7, 2016 at 16:16
  • $\begingroup$ May I know, how did you conclude $PH=(b/a)AH$ in the second paragraph, without invoking the fact coordinates of $P$ is $(a\cos\theta,b\sin\theta)$? $\endgroup$
    – Vishnu
    Nov 9, 2019 at 13:14
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    $\begingroup$ @Intellex Just consider similar triangles $ABP$ and $AOH$. $\endgroup$ Nov 9, 2019 at 14:45
  • $\begingroup$ @Aretino, Thank you. Could you please answer this follow-up question - Reasoning behind the $y$ coordinate of a point on an ellipse in the parametric form, if possible? $\endgroup$
    – Vishnu
    Nov 9, 2019 at 14:50

Why not simply

\begin{equation*} \left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=\cos ^{2}t+\sin ^{2}t=1 \end{equation*}

  • $\begingroup$ x may be cos(t). I don't see why you can say that (x/a) is cos(t) $\endgroup$ Sep 20, 2020 at 18:01

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