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Theorem: 4 normals can be drawn from any point to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. $( a > b > 0)$

I had a doubt in the proof of the above statement given in my book.

Proof given in Book. The normal at the point with eccentric angle $\phi$ is $ax\sec\phi - by \csc\phi = a^2 - b^2 $. Let P$(h,k)$ be any point in $\mathbb{R^2}$. If the normal passes through P: $$ah\sec\phi - bk \csc\phi = a^2 - b^2 = a^2e^2 $$ Let $ t = \tan \frac{\phi}{2}$, then $\cos \phi = \frac{1 - t^2}{1 + t^2}$ and $\sin \phi = \frac{2t}{1 + t^2}$. Substituting and rearranging, gives:

$$bkt^4 + 2t^3(ah + a^2e^2) + 2t(ah - a^2e^2) - bk = 0 $$

This is a quartic in $t$, hence has 4 roots, corresponding to 4 points on the ellipse which are conormal.

Doubt: My question is how do we know that the roots of that quartic equation are real and distinct? Because if not, then the theorem is not yet proved. I looked up the discriminant of a quartic on Wikipedia (https://en.wikipedia.org/wiki/Quartic_function), but the inequalities are too difficult to handle. How can I show that the roots are all real and distinct?

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marked as duplicate by Ng Chung Tak, Community Aug 30 '18 at 20:32

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  • $\begingroup$ What happens if your ellipse is a circle? $\endgroup$ – Andreas Blass Aug 30 '18 at 17:51
  • $\begingroup$ If it's a circle, then $a = b, e = 0$. Then two cases arise: $(h,k) = (0,0)$: Then the polynomial vanishes to zero, and is true for all $\phi$. $(h,k) \neq (0,0)$: The quartic becomes: $kt^4 + 2ht^3 + 2ht - k = 0 $. I don't know how to examine the nature of roots of this quartic too. $\endgroup$ – Kaind Aug 30 '18 at 18:03
  • $\begingroup$ Actually it's possible to factorize the quartic for the special case of a circle: $kt^4+2ht^3+2ht−k= (t^2 + 1) ( k t^2 + 2ht - k) = 0$. This always has 2 distinct real roots. $\endgroup$ – Kaind Aug 30 '18 at 18:08
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    $\begingroup$ One premise of the theorem is that a > b > 0, so the theorem expressly excludes the case of a circle. $\endgroup$ – Steve B Aug 30 '18 at 18:21
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    $\begingroup$ This theorem - as stated - is false, see here: mathpages.com/home/kmath505/kmath505.htm. Are you sure you wrote the exact statement? $\endgroup$ – Aretino Aug 30 '18 at 19:34
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I think the theorem is simply false. Here is a rendering of the set of lines normal to an ellipse $(a=1, b=1/2)$:

enter image description here

There is a region (if I had to guess, bounded by the evolute of the ellipse: it would be interesting to prove this) where every point is indeed intersected by four normals to the ellipse. Outside this region, only two normals can be drawn.

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  • $\begingroup$ I don't know man - the only reason I believed the theorem was true was because it was given in a standard book (S L Loney) and has been copied from there onto other texts too. As I have a doubt in the proof myself, I can't definitely say whether it's true or not? Either way, if it's wrong, could you correct it? $\endgroup$ – Kaind Aug 30 '18 at 18:43
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    $\begingroup$ Standard books can occasionally contain errors. That's why argument from authority is a logical fallacy. $\endgroup$ – Steve B Aug 30 '18 at 18:51
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    $\begingroup$ Comparing to this answer, the region is indeed bounded by the evolute. By that answer, the number of normals is $2$, $3$, or $4$, depending upon a point's relationship to that evolute. $\endgroup$ – Blue Aug 30 '18 at 20:10
  • $\begingroup$ @Blue: Yes that's the complete answer to the problem. Thank you! $\endgroup$ – Kaind Aug 30 '18 at 20:13
  • $\begingroup$ Here is an example of an interior point F where four normals to the ellipse exist. What you have to show is that since two normals always meet at a point, the locus of those points self-intersects allowing for 2 pairs of 2 points to meet at a single point. $\endgroup$ – ja72 Aug 30 '18 at 20:15
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Here is an example of such 4 normal:

enter image description here

Hope this helps!

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    $\begingroup$ This is two normals and two tangents. $\endgroup$ – ja72 Aug 30 '18 at 20:01
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    $\begingroup$ Those are 4 normal - 2 normal to the tangents drawn and one which is actually two merged. Slight move left or right will split the two through the center of the ellipse into two. Total 4. $\endgroup$ – Moti Aug 30 '18 at 20:07
  • $\begingroup$ This is what four normals through a single point F look like. $\endgroup$ – ja72 Aug 30 '18 at 20:13
  • $\begingroup$ This is the case when the point is in the ellipse. There is a range for such points in and out of the ellipse/circle. $\endgroup$ – Moti Aug 31 '18 at 0:52

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