Week 11, Day 1

Today we learned about hyperbolas, the set of all points (x,y) the difference of whose distances from two distinct fixed points (foci) is a positive constant. The definition of  hyperbola paralells that of an ellipse. The difference is that for an ellipse the sum of the distances between the foci and a point on the ellipse is foxed, whereas for a hyperbola the difference of these distances is fixed. Every hyperbola has two disconnected branches. The line through the two foci intersects a hyperbola at its two vertices. The line segment connecting the vertices is called the transverse axis, and the midpoint of the transverse axis is called the center of the hyperbola. The development of the standard form of the equation of a hyperbola is similar to that of an ellipse. The following are standard equations of a hyperbola, in which the center is at (h,k): 

The following diagrams illustrate the horizontal and vertical orientations of a hyperbola:

Example 1: Finding the Standard Equation of a Hyperbola

Find the standard form of the equation of the hyperbola with foci at (-1,2) and (5,2) and vertices at (0,2) and (4,2). By the Midpoint Formula, the center of the hyperbola occurs at the point (2,2). Furthermore, c = 3 and a = 2, and it follows that b^2 = c^2 - a^2 = 3^2 - 2^2 = 9 - 4 = 5. Thus, the equation of the hyperbola is [(x - 2)^2 / 4] - [(y - 2)^2 / 5] = 1. 

Asymptotes of a Hyperbola:

Each hyperbola has two asymptotes that intersect at the center of the hyperbola. The asymptotes pass through the vertices of a rectangle of dimensions 2a and 2b, with kts center at (h,k).

The asymptote for a horizontal transverse axis is: y = k +/- b/a(x - h). 

The asymptote for a vertical transverse axis is: y = k +/- a/b(x - h). 

Example 2: Finding the Asymptotes of a Hyperbola

Sketch the hyperbola given by 4x^2 - 3y^2 + 8x + 16 = 0 and find the equations of its asymptotes. 

4x^2 - 3y^2 + 8x + 16 = 0

4(x^2 + 2x) - 3y^2 = -16

4(x^2 + 2x + 1) - 3y^2 = -16 + 4

4(x + 1)^2 - 3y^2 = -12

[y^2 / 4] - [(x + 1)^2 / 3] = 1

From this equation you can conclude that the hyperbola is centered at (-1,0), has vertices at (-1,2) and (-1,-2), and the ends of the conjugate axis occur at (-1 - (3)^1/2, 0) and (-1 + (3)^1/2, 0). To sketch the hyperbola, draw a rectangle through these four points. The asymptotes are the lines passing through the corners of the rectangle. Finally, using a = 2 and b = (3)^1/2, you can conclude that the equations of the asymptotes are: y = [2/(3)^1/2] (x + 1)] and y = [-2/(3)^1/2] (x + 1)]. 

Example 3: Using Asymptotes to Find the Standard Equation

Find the standard form of the equation of the hyperbola having vertices at (3,-5) and (3,1) and with asymptotes y = 2x - 8 and y = -2x + 4. By the Midpoint Formula, the center of the hyperbola is at (3,-2). Furthermore, the hyperbola has a vertical transverse axis with a = 3. From the given equations, you can determine the slopes of the asymptotes to be: m(1) = 2 = a/b and m(2) = -a/b and because a = 3, you can conclude that b = 3/2. Thus, the standard equation is [(y + 2)^2] / 9 - [(x - 3)^2] / 9/4 = 1. 

Eccentricity:

As with ellipsese, the eccentricity of a hyperbola is e = c / a and because c > a, it follows that e > 1. If the eccentricity is large, the branches of the hyperbola are nearly flat. If the eccentricity is close to 1, the branches of the hyperbola are more pointed. 

Classifying a Conic from its General Equation:

The graph of Ax^2 + Cy^2 + Dx + Ey + F = 0 is one of the following.

1. Circle: A = C

2. Parabola: AC = 0 (A = 0 or C = 0, but not both)

3. Ellipse: AC > 0 (A and C have like signs)

4. Hyperbola: AC < 0 (A and C have unlike signs)