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I am posting the following question under homework category. I hope I will have very good answer from mathematicians about conic sections.

I have seen closely the conic sections and their properties. I have seen this result in hyperbola. In this I would like to increase eccentricity (e) of hyperbola gradually. i.e., if eccentricity of hyperbola is large, then hyperbola becomes a straight line. How and why? Could you explain please?

Also, I would like to know if we can apply the same for the ellipse as well as the parabola. If yes, what result can we obtain?

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What is the actual homework problem you're trying to solve? –  Henning Makholm Nov 2 '11 at 1:33

2 Answers 2

The hyperbola does not really "become a line if the eccentricity is large".

First, let's clarify. Take the hyperbola given by $$x^2 - \frac{y^2}{b^2} = 1.$$

The eccentricity of this hyperbola is $$\varepsilon =\sqrt{1 + b^2},$$ and we can ask what happens to the hyperbola as $b\to\infty$; that is, as $b$ (and the eccentricity with it) grows larger.

For any particular value of $b$, no matter how large, we will still have a hyperbola.

What we may be interested in, though, is what happens "at the limit". This is like going to the projective plane (where we have added a "line at infinity") and moving the foci of the hyperbola further and further away from the origin. "At the limit", we have made both foci of the hyperbola coincide, and be at the line at infinity. What we obtain is a degenerate hyperbola, which is given by the equation $$x^2 = 1.$$ But $x^2 = 1$ is equivalent to $(x-1)(x+1) = 0$, which gives you "two vertical lines". This is a degenerate case, and we don't actually "reach it" at any particular value of $b$. It is the "limiting case".

A similar thing happens with the other conics.

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Good explanation. Thank you soooo much –  mathew Nov 2 '11 at 17:58

Also, I would like to know, can we apply the same for ellipse as well as parabola. If yes, what result we can obtain?

Just to get it out of the way: eccentricities are positive quantities, since they are defined as ratios of distances, which are positive quantities in themselves. Arturo has explained the hyperbolic case, so I shall confine myself to the other cases.

By definition, a parabola has unit eccentricity, so one cannot really speak of "parabolas of increasing eccentricity". In fact, parabolas are like circles, in that all parabolas are similar (differing only in scale).

For ellipses, the eccentricity $\varepsilon$ is constrained to lie in the interval $(0,1)$; one can thus study the behavior as $\varepsilon\to 0$ or $\varepsilon\to 1$.

To study the behavior of an ellipse as $\varepsilon\to 0$, consider the standard form for a central conic with center at the origin and axes lying on the coordinate axes:

$$\frac{x^2}{a^2}+\frac{y^2}{a^2(1-\varepsilon^2)}=1$$

where $a$ is the length of the principal axis. Taking the limit as $\varepsilon\to 0$ yields

$$\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$$

and we have a circle of radius $a$. Geometrically, as the eccentricity shrinks, the two foci come closer and closer, up until they coincide with the center in the limit. (In this sense, a circle is a degenerate ellipse!)

As $\varepsilon\to 1$, to take the other extreme, if one fixes the position of one focus, the other focus will eventually shoot into the wild blue yonder (go to infinity), eventually leading to the parabola when $\varepsilon=1$.

To see this quantitatively, consider the general equation of a conic with one focus at the origin and principal axis coinciding with the horizontal axis:

$$(1-\varepsilon^2)x^2+y^2-2\varepsilon p x=p^2$$

where $p$ is the semilatus rectum. When $\varepsilon=1$, we have a parabola as expected. To see the behavior of an ellipse as $\varepsilon\to 1$, we treat the general equation as a generic central conic; we find that the coordinates of the "other focus" are

$$\left(\frac{2\varepsilon p}{1-\varepsilon^2},0\right)$$

and it is apparent that the other focus does "shoot off to infinity" as $\varepsilon\to 1$.

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Very interesting in explanation. Once again thank you so much... –  mathew Nov 2 '11 at 17:58

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