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HYPERBOLAS
- A set of all points in the xy-plane where the difference of the distances from two fixed points is a positive constant. The fixed points are called foci.
- The vertices of a hyperbola are the points where the verse axis of a hyperbola intersects the transverse axis. The transverse axis is the line segment joining the intercepts. The conjugate axis is a line segment that passes through the center of the hyperbola and is perpendicular to the transverse axis. The midpoint of the transverse and conjugate axis is the center of the hyperbola. The asymptotes are the lines that approaches a given curve arbitrarily closely, however, never touching. HYPERBOLA MATRIX HORIZONTAL VERTICAL Equations Standard If C is at origin (0, 0) Foci Vertices Asympto tes If C (h, k) Foci Vertices Asympto tes Conditions Transver se Axis Horizontal Vertical Conjugat e Axis Vertical Horizontal Foci Left and right of C Above and below C Vertices Left and right of C Above and below C Constant values (ex. If foci has same h, it’s vertical) Foci k h Vertices k h Other relations Auxiliary circle = Exercise 1. A. Graph the hyperbola. Determine the foci, vertices, and asymptotes.
Sol: When the equation starts with positive , then that equation denotes a horizontal ellipse whose branches open to the left and to the right.
Sol: When the equation starts with positive , then that equation denotes a vertical ellipse whose branches open upwards and downwards.
Sol: Multiply the reciprocal of the right side of the equation to get 1.
Sol:
Sol: Multiply the reciprocal of the right side of the equation to get 1. Cancelling out by division is sometimes confusing, just like in this example wherein at first glance, 18 and 32 could be 9 and 16.
Sol: Get the PST and multiply once again the right’s reciprocal to the equation.
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19.Foci (-3, 1) and (7, 1), transverse axis of length 4 units Sol: We know that the length of the transverse axis is 2a. 20.Vertices (-3, 0) and (3, 0), through Sol: Since there are no foci given to use the formula , use the original standard form and substitute x and y to get b. 21.Foci (0, -4) and (0, 4), though (-3, 5) Sol: Use the formula to get a. 22.Center (0, 0), vertical transverse axis, through and Sol: Substitute the two points in two different equations and make a system. For simplicity’s sake, let A and B take the place of. To get A, use the elimination method. Multiply the first equation by a product that will result into the cancellation of B. In this case, multiply the first equation by to get. Since both are technically variations of the same hyperbola, we can combine them to get the answer. Get B by substituting A into an equation.
23.Asymptotes and , horizontal transverse axis of length 12 units Sol: If the transverse axis is horizontal, then it means that the hyperbola is horizontal. Thus, the formula for asymptotes is
. Thus, we get the ratio of a and b. Make a system to get h and k. When expanded, the asymptotes are and. However, when minimized, they take the form of where. Thus, here, our c values are for the first asymptote and for the second. 24.Asymptotes and , vertical transverse axis of length 12 units Sol: If the transverse axis is vertical, then it means that the hyperbola is vertical. Thus, the formula for asymptotes is . Thus, we get the ratio of a and b. Remember that the center is the intersection of the two asymptotes. Therefore, if the asymptotes of two equations are the same, then they have the same center. Refer to the previous number. 25.Asymptotes , focus (9, 3) Sol: Cases when only the asymptote and a focus is given, we can find out the coordinates of our center and the kind of hyperbola we are facing. Since our center and focus are constant k, we are facing a horizontal hyperbola. Since is a ratio that can be simplified, we assume that a is not simply a and b is not simply b. They are possibly multiplied
Sol: This is a hyperbola. A circle must have the same coefficient for and. A parabola can only have one variable, either x or y, squared. An ellipse would have different coefficients for and , with both being positive. The hyperbola requires the coefficient of one squared variable, either x or y, to be negative. Thus, this is a hyperbola.
Sol: This is a parabola. A circle must have the same coefficient for and. The hyperbola requires the coefficient of one squared variable, either x or y, to be negative. An ellipse would have different coefficients for and , with both being positive. A parabola can only have one variable, either x or y, squared. Thus, this is a parabola.
Sol: This is a hyperbola.
Sol: This is an ellipse.
Sol: This is a parabola. D. Solve each of the following problems. 35.Two lighthouses, 1200 m apart on a shore that runs west to east, are at the points F 1 (-600, 0) and F 2 (600,0). Here, the x-axis is super-imposed on this straight shore. Both lighthouses, at the same moment, send a message out to a ship at the sea. The ship received the message from F 2 3. seconds before the message from F 1. Assume that the messages, sent as radio signals, traveled at a constant speed of Sol: Convert first microseconds into seconds so that we’re working with the same unit. In this case where there are no given a or b, remember that. We do not have the coordinate of the ship so we cannot substitute anything in P. However, we use the difference in time for it also partially represents the difference in distance. Remember that d = rt. a. Find the equation of the hyperbola comprising the possible locations of the ship. Which branch is it, specifically? Sol:
Since the ship received the message from F 2 earlier, we are assuming that the ship is within the right branch of the hyperbola. b. If the ship is directly north of F 2 , how far is it from the shore? Sol: Substitute the x-coordinate of F 2 in the equation to get its distance from the shore. c. If the ship is 400 m away from the shore, what are its coordinates? Sol: Replace y with 400. 36.A ship out sea sent a message to two control towers by a straight shoreline, at the points F 1 (-3, 0) and F 2 (3,
- (all units in km), with the x-axis running along the shore. The station at F 2 received the message 8 seconds after the message was received at F 1. Assume that the message traveled at 0.33 km/s. a. Find the equation of the hyperbola comprising the possible locations of the ship. Which branch is it, specifically? Sol: If F 2 received the message from the ship 8 seconds after F 1 received it, it would mean that F 1 is closer to the ship for it received the
40.When a comet follows a parabolic or hyperbolic (one branch) path with the sun at its focus, it only visits once and never returns. It is closest to the sun when it is at the vertex. Suppose the sun is at (100, 0), and a comet is closest to the sun when it is at (70, 0). Suppose the units are in 1,000, km. a. Find the equation of its path if it is a parabola. Sol: b. Find the equation of its path if it is a hyperbola centered at the origin. Sol: 41.Find the equation of the hyperbola, whose vertices are the foci of the ellipse and whose foci are the vertices of this ellipse. Sol: A hyperbola is basically an inverted/reversed ellipse. The values and their relations are only switched around. In a hyperbola, an ellipse’s c is a and an ellipse’s a is c. However, b does not change. 42.Find the equation of the hyperbola having vertices (-5, 4) and (1, 4), and having a focus on the line. Sol: We know that this is a horizontal parabola because the vertices are constant at k. Find the center. Then, replace y with 4 to
know the coordinate of the focus located on that line. 43.Find the equation of the ellipse that is tangent to all vertices of the hyperbolas and Sol: There are two hyperbolas here: one that is vertical and hence will have a vertical transverse axis, and one that is horizontal that will have a horizontal transverse axis. The length of these axes will be identical to the lengths of the ellipse’s major and minor axes. Assess which axis is larger. That will be the major axis. 44.Consider the equation , where r is some constant not equal to and -2. Find the value(s) of r so that the graph is: a. A circle Sol: An equation can only be a circle if you can cancel out its coefficients and denominators. Therefore, the denominators/coefficients of a circle should be equal. b. An ellipse with a horizontal major axis Sol: An ellipse with a horizontal major axis takes on the form . c. An ellipse with a vertical major axis Sol: An ellipse with a horizontal major axis takes on the form .
Get the distance between F(1,3) and y = - 4 first. Remember the relationship between point, directrix, and focus. Let y = - 4 be the directrix. b. Find the equation of the set of all points P (x, y) for which PF =
- What conic section is this? Sol: Find the eccentricity of the equation. The eccentricity is the unique shape of the curve of a conic section. Circle has eccentricity of 0. Ellipse has eccentricity of > 0 but < 1. Parabola has eccentricity of 1. Hyperbola was eccentricity of > 1. To get the equation of the hyperbola, equate. This conic section is a hyperbola. c. Find the equation of the set of all points P (x, y) for which PF =
. What conic section is this? Sol: This conic section is an ellipse. 47.The set of all points P (x, y), such that the difference of its distances from the two points and is , is a hyperbola, shown in the figure below. The points are the foci and the center of the hyperbola is the origin (midpoint of the foci). Find the equation of this hyperbola. Sol:
Since , we know that 2a is
. Find a and the coordinates of the vertices. Since the transverse axis of the hyperbola lies on the line y = x, substitute y to x when getting the vertices. _This reviewer is for educational purposes only. This pdf is not free from errors. Please correct on your own if you see them or inform me if you want a new copy. I tried my best in proofreading this document but I may have missed some things. Please do not redistribute or share. If someone you know wants a copy, refer them to me. Thanks!
- Bernice Danielle E. Castillo 12STEM-_
