How to know if a circle/ellipse is a point or empty space from the general equation? And hyperbola as lines?

Question

I need a quick and simple way to identify from the general equation if a circle or ellipse is a point or empty space and also the same for the hyperbola but whether the equation represents a set of intersecting lines.

Answer 1

What are you using for the general equation? Are you using the following?

Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

This is the most general equation for an ellipse or hyperbola. The major and minor axes can be rotated w.r.t. the x and y axes, and the center of the conic section can be translated from the x-y origin. In this case, you will have to cast the equation into a symmetric matrix, find the eigenvalues and eigenvectors to determine the shape, orientation, and center of the conic section.

If your "general" equation is actually for a conic section aligned and centered to the coordinate axes, then you have a simpler problem.

Ellipse: (x^2 / a^2) + (y^2 / b^2) = 1

As 'a' and 'b" approach 0, the ellipse collapses to a point.

East-west hyperbola: (x^2 / a^2) - (y^2 / b^2) = 1

As 'a' and 'b" approach 0, the hyperbola approaches two intersecting lines.

Answer 2

The general equation for a quadratic can be written as:

ax² + bxy + cy² + dx + ey + f = 0

To determine quickly what kind of curve this represents we often use a "discriminant".

4ac - b² > 0 ellipse
4ac - b² = 0 parabola
4ac - b² < 0 hyperbola

However this does not tell us if it a regular conic section or a degenerate case. For that we need to take a closer look.

Unfortunately there is not complete agreement with respect to notation for the general quadratic. Sometimes the general equation has some "2"s in it as follows.

ax² + 2bxy + cy² + 2dx + 2ey + f = 0

The "2"s allow us to avoid fractions in examining the discriminant, but it also changes the discriminant.

ac - b² > 0 ellipse
ac - b² = 0 parabola
ac - b² < 0 hyperbola

Again, this does not tell us if it a regular conic section or a degenerate case. To find out take a closer look.

The following link tells you all the cases for regular or degenerate cases of ellipses, parabolas, and hyperbolas.

http://mathforum.org/dr.math/faq/formulas/faq.analygeom_2.html#twogenquads

As for whether or not an ellipse is also a circle. First test to see if it is an ellipse. If it is and a = c, then it is also a circle.

Answer 3

THE ALGEBRAIC approach ok there are 2 procedures you are able to come to a decision this situation, with calculus, or the headcracking algebra. i will use the algebraic approach, to exhibit it would want to certainly be carried out. ok, so your parabola is xy = 3 or y = 3/x. and also you element is (4,3/4) Any line crossing (4,3/4) has the type (3/4)= m(4) + b from (y = mx+b), in spite of tangency. Your interest now, is to locate the slope m. separate m from (3/4)= m(4) + b, and look at were given m = [(3/4) - b] / 4 ok, now you've the slope of the line in words of b, now locate the slope of the hyperbola. We in basic terms reported that xy = 3 is likewise y = 3/x, yet y = mx + b too. Plug contained in the y. mx + b = 3/x come to a decision for m... m = [(3/x)-b] / x Now you've 2 m's. Set them equivalent to at least one yet another and are available to a decision for an x. [(3/4) - b] / 4 = [(3/x)-b] / x go multiply and distribute... (3/4)x - xb = (12/x) - 4b bypass each little thing to the left... (3/4)x - xb - (12/x) + 4b = 0 shall we cases each little thing by 4x to get rid of that pesky denominator... 3x^2 - 4x^2b - 40 8 + 16xb = 0 rearrange somewhat... 3x^2 - 40 8 - 4x^2b + 16xb = 0 component out some words... 3(x^2 - 16) - 4xb (x + 4) = 0 strengthen (x^2 - 16) and remultiply to get (x + 4) on my own. 3x -12(x + 4) - 4xb (x + 4) = 0 by guidelines of factoring...you are able to combine those to type... (3x - 12 - 4xb)(x+4) = 0 x is now equivalent to -4 from the 2d time period, plug -4 as x into (3x - 12 - 4xb) = 0 and are available to a decision for b. (3(-4) - 12 - 4(-4)b) = 0 -12 - 12 + 16b = 0 -24 + 16b = 0 b = 24/16 or 3/2. Phew...that grow to be lengthy, ok its about time, you stumbled on b, plug that into (3/4)= m(4) + b to locate m. (3/4)= 4m + (3/2) m = -3/16 ultimately, you are able to make YOUR EQUATION utilising m and b. y = (-3x/16) + (3/2) <==== that's YOUR TANLINE. to locate the traditional...in basic terms use the unfavorable reciprocal of the slope... it quite is 16/3, and plug it into y - y1 = m(x - x1) y - (3/4) = 16/3(x - 4) y = (16x/3) - (sixty 4/3) + (3/4) <=== that's YOUR time-honored. solid luck, i'm hoping that's useful. again, you should use calculus and save your self an vast era of time by in basic terms utilising the by-product of the hyperbola.