Instructions:Answer all the following questions in the space provided. Simplify all answers.
- Identify the centre, opening direction, vertices and the slopes of the asymptotes for the hyperbola defined by:
a) x2/ 49 - y2/ 16 = 1
b) -x2/ 4 + (y + 4)2/ 64 = 1
c) (x + 8)2/ 9 - (y - 16)2/ 8 = 1
d) -(x + 6)2/ 25 + (y - 12)2/ 7 = 1
| | a | b | c | d |
Centre
| (0, 0) | (0, -4) | (-8, 16) | (-6, 12) |
| Opening Direction | left and right | up and down | left and right | up and down |
| Vertices | (7, 0) and (-7, 0) | (0, 4) and (0, -12) | (-5, -16) and (-11, 16) | (-6, 12 + Ö7) and (-6, 12 - Ö7) |
| Slopes of Asymptotes | ±(4/7) | ±4 | ±(2Ö2)/3 | ±(Ö7)/5 |
- A hyperbola centred at the origin opens up and down and has asymptotes with slopes of +3/2 and -3/2.
| a) Write the equation of this hyperbola in standard form. |
-x2/4 + y2/9 = 1 |
b) Write the equation of this hyperbola in general form. |
-9x2 + 4y2 - 36 = 0 |
c) Sketch the graph of this hyperbola.
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- A hyperbola centred at (3, 1) opens left and right and has asymptotes with slopes of +5 and -5.
| a) Write the equation of this hyperbola in standard form. |
(x - 3)2 - (y - 1)2/25 = 1 |
b) Write the equation of this hyperbola in general form. |
25x2 - y2 - 150x + 2y + 199 = 0 |
c) Sketch the graph of this hyperbola.
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- Describe the effect that varying h and k in the standard equations (x - h)2/ a2 - (y - k)2/ b2 = 1 and -(x - h)2/ a2 + (y - k)2/ b2 = 1 has on the graph of a hyperbola by completing the following chart.
The Effect of h and k on the graphs of (x - h)2/ a2 - (y - k)2/ b2 = 1 and -(x - h)2/ a2 + (y - k)2/ b2 = 1
| Variable | The value of the variable decreases | The value of the variable increases | The value of the variable is 0. |
|---|
| h | the hyperbola shifts to the left | the hyperbola shifts to the right | the vertices of the hyperbola will always be on the y-axis |
| k | the hyperbola shifts down | the hyperbola shifts up | the vertices of the hyperbola will always be on the x-axis |
- Describe the effect that varying a and b in the standard equations (x - h)2/ a2 - (y - k)2/ b2 = 1 and -(x - h)2/ a2 + (y - k)2/ b2 = 1 has on the graph of a hyperbola.
The variables a and b in the equation (x - h)2/ a2 - (y - k)2/ b2 = 1 or -(x - h)2/ a2 + (y - k)2/ b2 = 1 affect the location of the hyperbola's vertices and the slope of it's asymptotes.
- The equation -9x2 + y2 + 144x - 6y - 648 = 0 defines a hyperbola. Identify the centre, opening direction, vertices and the slopes of the asymptotes of this hyperbola.
| Centre: (8, 3) | | Vertices: (8, 12) and (8, -6) |
Opening direction: up and down | |
Slopes of Asymptotes: ±3 |
- Sketch the graph of the hyperbola defined by the equation 64x2 - 100y2 - 6400 = 0.
- If the equation Ax2 + By2 + Cx + Dy + F = 0 and the equation Ax2 + By2 = P define a hyperbola, then what must be true about the values of the coefficients A and B in both equations?
If the equation Ax2 + By2 + Cx + Dy + F = 0 and the equation Ax2 + By2 = P define a hyperbola, then A and B must be different signs. That is one must be positive and one must be negative.
- A hyperbola is defined by the standard equation (x - 6)2 - (y - 6)2 = 1 and by the general equation x2 - y2 - 12x + 12y - 1 = 0.
a) Show that these equations are equivalent.
One way is to expand the standard equation:
(x - 6)2 - (y - 6)2 = 1
x2 - 12x + 36 - (y2 - 12y + 36) = 1
x2 - 12x + 36 - y2 + 12y - 36 = 1
x2 - y2 - 12x + 12y - 1 = 0
This expanded equation is exactly equal to the general equation given in the question, therefore, we have just shown that the two equations given are equivalent.
Another way to show equivalency would be to complete the square on the general equation:
x2 - y2 - 12x + 12y - 1 = 0
(x2 - 12x) - (y2 - 12y) - 1 = 0
(x2 - 12x + 36) - 36 - (y2 - 12y + 36) - (-36) - 1 = 0
(x - 6)2 - (y - 6)2 = 1 + 36 - 36
(x - 6)2 - (y - 6)2 = 1
This completed equation is exactly equal to the standard equation given in the question, therefore we have just shown that the two equations given are equivalent.
b) When two equations are equivalent they have identical solutions sets. Verify that the point (5, 6) is a solution of both equations.
Substitute the point (5, 6) into both equations:
(x - 6)2 - (y - 6)2 = 1
(x - 6)2 - (y - 6)2
= (5 - 6)2 - (6 - 6)2
= (-1)2 - 0
= 1
x2 - y2 - 12x + 12y - 1 = 0
x2 - y2 - 12x + 12y - 1
= (5)2 - (6)2 - 12(5) + 12(6) - 1
= 25 - 36 - 60 + 72 - 1
= 0
- A hyperbola is formed when a plane cuts both naps of a double-napped cone. Describe what happens to the hyperbola as:
a) The plane moves closer to the vertical axis.
The vertices of the hyperbola get closer together.
b) The plane moves farther away from the vertical axis.
The vertices of the hyperbola get farther apart.
c) The plane cuts exactly through the vertical axis.
The result is an intersecting pair of lines.
- Explain why (or show how) the horizontal and vertical axes of symmetry can be used to describe the slopes of the asymptotes as +b/a and -b/a for a hyperbola defined by (x - h)2/ a2 - (y - k)2/ b2 = 1.
A hyperbola defined by the equation (x - h)2/ a2 - (y - k)2/ b2 = 1 has vertices on the horizontal axis of symmetry. For this hyperbola we can label the following distances and points on the axes of symmetry:
the distance between the vertices is 2a
the distance between the centre of the hyperbola and each vertex is |a|
there exists 2 points on the vertical axis such that the distance between each point and the centre of the hyperbola is |b|
Using these distances and points, a rectangle of width 2a and height 2b can be drawn in the graph of a hyperbola. The asymptotes of the hyperbola intersect the corners of this rectangle. Now the rectangle can be used to find the slopes of the asymptotes. Using the definition of slope, the slopes of the asymptotes are ± b/a.
A similar explaination can be used for the equation -(x - h)2/ a2 + (y - k)2/ b2 = 1.
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