matematika lesson
hyperbola
ne cone if sliced horizontally, then the cut circle. If the cone is cut at an angle (and not cut the base), then formed an ellipse. If the slices cut its base and cut it vertically, then formed a hyperbole. If the slices are not cut off its base and cut it vertically, then formed a parabola.
We know it is the ellipse equation
\ frac {x ^ 2} {a ^ 2} + \ frac {y ^ 2} {b ^ 2} = 1
Hyperbola equation similar to equation ellipse. It's just that the sign is not positive, but negative. Hyperbolic equation is as follows:
\ frac {x ^ 2} {a ^ 2} - \ frac {y ^ 2} {b ^ 2} = 1
How graphic image of a hyperbole?
For instance the image graph of the equation: \ frac {x ^ 2} {16} + \ frac {y ^ 2} {9} = 1
Do have any idea to connect the equation with its graph images?
When y = 0, then \ frac {x ^ 2} {16} = 1 so x = \ pm 4
\ Frac {x ^ 2} {a ^ 2} - \ frac {y ^ 2} {b ^ 2} = 1
When y = 0, then x = \ pm a, a what we call a peak
What is the role of b?When we write the equation of a hyperbola in x, then we can menulsikan\ Frac {x ^ 2} {a ^ 2} - 1 = \ frac {y ^ 2} {b ^ 2}\ Frac {(x ^ 2-a ^ 2) (b ^ 2)} {a ^ 2} = y ^ 2\ Pm \ sqrt {\ frac {(x ^ 2-a ^ 2) (b ^ 2)} {a ^ 2}} = yy = \ pm \ frac {b} {a} \ sqrt {x ^ 2-a ^ 2}
For large values of x, \ sqrt {x ^ 2-a ^ 2} behaves like x, ie if x \ to \ infty then \ sqrt {x ^ 2-a ^ 2}-x \ to 0. So y is suchy = \ frac {b} {a} x or y = - \ frac {b} {a} xThe two lines are asymptotes of the graph hyperbola equation.
We've got the b (see the picture), note the triangle with sides a, b and c in the figure. We get c ^ 2 = a ^ 2 + b ^ 2, the coordinates of the focal point (c, 0)
PARABOLA FOR VERTICAL, READERS CAN ADJUST(There is one about hyperbole vertical, number 3)
PROBLEM:1. Determine both the focal point of hyperbole: \ frac {x ^ 2} {16} - \ frac {y ^ 2} {9} = 1Answer:\ Frac {x ^ 2} {a ^ 2} - \ frac {y ^ 2} {b ^ 2} = 1 if we look at the general, then we get a = 4 and b = 3. Of course we are looking at the formula c c ^ 2 = a ^ 2 + b ^ 2, and we get c = 5.So the coordinates of the focal point of the hyperbola is (PM 5.0)
2. Determine the line asymptotes of the hyperbola: \ frac {x ^ 2} {16} - \ frac {y ^ 2} {9} = 1Answer:\ Frac {x ^ 2} {a ^ 2} - \ frac {y ^ 2} {b ^ 2} = 1 if we look at the general, then we get a = 4 and b = 3. Both of us know as the asymptote y = \ pm \ frac {b} {a} x, then we get the second asymptote is y = \ pm \ frac {3} {4} x
3. Problem for vertical hyperbole.Determine both the culmination, the focal point and the line asymptotes for the hyperbola: \ frac {y ^ 2} {16} - \ frac {x ^ 2} {9} = 1atau can also be written: - \ frac {x ^ 2} {9 +} \ frac {y ^ 2} {16} = 1Answer:When we take y = 0, we may not be able to find the value of x. because the form - \ frac {x ^ 2} {9} = 1 is not going to be fulfilled for any x.We take x = 0, we get y = 4. This is the peak. (Pictures only scratch-marks on the x = 4 and x = -4 as a peak, and then drawing simple hyperbole)
Note that we use the general equation: \ frac {x ^ 2} {a ^ 2} - \ frac {y ^ 2} {b ^ 2} = 1(A were his x, is under (the denominator) of x and y b was hers, was under (the denominator) of y)Thus, for the matter of: - \ frac {x ^ 2} {9} + \ frac {y ^ 2} {16} = 1We get a = 3 and b = 4So that was the line asymptote is y = \ pm \ frac {4} {3} xTo find the focal point, we need to find c, we get that c is equal to 5. Because vertical hyperbole, the coordinates of the point c is (0, \ pm c) is equal to (0, \ pm 5)
When y = 0, then x = \ pm a, a what we call a peak
What is the role of b?When we write the equation of a hyperbola in x, then we can menulsikan\ Frac {x ^ 2} {a ^ 2} - 1 = \ frac {y ^ 2} {b ^ 2}\ Frac {(x ^ 2-a ^ 2) (b ^ 2)} {a ^ 2} = y ^ 2\ Pm \ sqrt {\ frac {(x ^ 2-a ^ 2) (b ^ 2)} {a ^ 2}} = yy = \ pm \ frac {b} {a} \ sqrt {x ^ 2-a ^ 2}
For large values of x, \ sqrt {x ^ 2-a ^ 2} behaves like x, ie if x \ to \ infty then \ sqrt {x ^ 2-a ^ 2}-x \ to 0. So y is suchy = \ frac {b} {a} x or y = - \ frac {b} {a} xThe two lines are asymptotes of the graph hyperbola equation.
We've got the b (see the picture), note the triangle with sides a, b and c in the figure. We get c ^ 2 = a ^ 2 + b ^ 2, the coordinates of the focal point (c, 0)
PARABOLA FOR VERTICAL, READERS CAN ADJUST(There is one about hyperbole vertical, number 3)
PROBLEM:1. Determine both the focal point of hyperbole: \ frac {x ^ 2} {16} - \ frac {y ^ 2} {9} = 1Answer:\ Frac {x ^ 2} {a ^ 2} - \ frac {y ^ 2} {b ^ 2} = 1 if we look at the general, then we get a = 4 and b = 3. Of course we are looking at the formula c c ^ 2 = a ^ 2 + b ^ 2, and we get c = 5.So the coordinates of the focal point of the hyperbola is (PM 5.0)
2. Determine the line asymptotes of the hyperbola: \ frac {x ^ 2} {16} - \ frac {y ^ 2} {9} = 1Answer:\ Frac {x ^ 2} {a ^ 2} - \ frac {y ^ 2} {b ^ 2} = 1 if we look at the general, then we get a = 4 and b = 3. Both of us know as the asymptote y = \ pm \ frac {b} {a} x, then we get the second asymptote is y = \ pm \ frac {3} {4} x
3. Problem for vertical hyperbole.Determine both the culmination, the focal point and the line asymptotes for the hyperbola: \ frac {y ^ 2} {16} - \ frac {x ^ 2} {9} = 1atau can also be written: - \ frac {x ^ 2} {9 +} \ frac {y ^ 2} {16} = 1Answer:When we take y = 0, we may not be able to find the value of x. because the form - \ frac {x ^ 2} {9} = 1 is not going to be fulfilled for any x.We take x = 0, we get y = 4. This is the peak. (Pictures only scratch-marks on the x = 4 and x = -4 as a peak, and then drawing simple hyperbole)
Note that we use the general equation: \ frac {x ^ 2} {a ^ 2} - \ frac {y ^ 2} {b ^ 2} = 1(A were his x, is under (the denominator) of x and y b was hers, was under (the denominator) of y)Thus, for the matter of: - \ frac {x ^ 2} {9} + \ frac {y ^ 2} {16} = 1We get a = 3 and b = 4So that was the line asymptote is y = \ pm \ frac {4} {3} xTo find the focal point, we need to find c, we get that c is equal to 5. Because vertical hyperbole, the coordinates of the point c is (0, \ pm c) is equal to (0, \ pm 5)
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