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ELLIPSES
- A set of all points in the xy-plane where the sum of the distances from two fixed points is a positive constant. The fixed points are called foci.
- The closes point from the foci to the ellipse are the vertices. The distance from the center to the closes point of the ellipse are the covertices. The line that contains the center and foci is called the principal (major) axis. The line that contains the center and the covertices is called the minor axis. ELLIPSE MATRIX HORIZONTAL VERTICAL Equations Standard where a > b where a > b If C is at origin (0, 0) Foci Vertices Covertice s If C (h, k) Foci Vertices Covertice s Conditions Major Axis Horizontal Vertical Minor Axis Vertical Horizontal Foci Left and right of C Above and below C Vertices Left and right of C Above and below C Covertice s Above and below C Left and right of C Constant values (ex. If foci has same h, it’s vertical) Foci k h Vertices k h Covertice s h k Other relations Exercise 1. A. Graph the ellipse and determine the foci.
- Sol: In order to see whether the ellipse is vertical or horizontal, look for the position of. Since a > b, a can either be found on the first and second term of the equation. If it is found on the first term, it is a horizontal ellipse. If found on the second term, it is a vertical ellipse. If horizontal,. If vertical,.
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Sol: There are cases where instead of seeing the a and b values in the denominator, we see them together with the x and y variables. It goes without saying that these values should be canceled out. In such cases, prioritize the right side of the equation more than the left. Multiply both sides by the reciprocal of the value on the right to get 1 on the right side and cancel out the values with the variables on the left side.
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Sol: If C(h, k) is given, that means that the value of the foci, vertices, and covertices do not solely rely on the values of a, b, and c. Analyze first if the ellipse is horizontal or vertical. If horizontal, . If vertical,.
To answer this equation, we must get the PST (Perfect Square Trinomial) of variables x and y.
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Sol: B. Find the equation in standard form of the ellipse with the specified features.
32.Center (-2, 4), major axis length of 12, vertical minor axis length of 6 Sol: 33.Minor axis length of 8, foci 5 units below and above the center (-2, 4) Sol:
34.Major axis length of 4 √ 13 , foci 4 units
below and above the center (-2, 4) Sol: 35.Vertices (-5, -2) and (11, -2), minor axis length of 12 Sol: 36.Vertices (3, -14) and (3, 6), a focus at (3, 2) Sol: 37.Covertices (2, 2) and (2, -6), a focus at (-5, -2) Sol: 38.Foci (5, -8) and (5, 2), a vertex 3 units above a focus Sol:
39.Foci (-1, 5) and (9, 5), a vertex 4 units to the left of a focus Sol: 40.Foci (-1, 4) and (15, 4), a covertex 10 units away from a focus Sol: In the case where a covertex is given instead of a vertex, remember that a covertex can also be a point P (x, y) where. However, since the covertices are placed at the minor axis which splits the ellipse directly in half, the distance between a focus and a covertex is simply a or. 41.Foci (-4, -4) and (-4, 10), a covertex
6 √ 2 units away from a focus
Sol: 42.Covertex (4, -11), focus (-3, -5), horizontal major axis Sol: Vertices, foci, and the center all have the same k if horizontal. The center and covertices have the same h. The distance between a focus and a covertex is a.
43.Covertex (-4, √ 94 ), focus (0, -11),
horizontal minor axis Sol: Vertices, foci, and the center all have the same h if vertical. The center and covertices have the same k.
46.Covertices (-3, -7) and (-3, 7), through
Sol: 47.A vertex (18, -2), a focus (12, -2), minor axis of length 24 Sol: The solutions to this problem is relatively simple and hinges foremost on basic algebra logic and substitution. Assume that the vertex and the foci lay on the same side of the ellipse. We know that b=12 because the length of the minor axis is 2b. Thus, to get a and c, we will use the equation. We know that V(h+a, k) and F(h+c, k). From there we derive that h+a=18 and h+c=12. Solve by getting two different equations and substituting one into the other. I have colored the key equations red for you to be able to follow the procedure. 48.A vertex (-7, -10), a focus (-7, 8), minor axis length of 24 Sol: Assume that the vertex and the focus are not on the same side of the ellipse. C. Solve the following problems. 49.A basketball court is inside a gymnasium whose ceiling is in semi- elliptical shape. The two rings are 90 ft
away from each other and standing at the foci. If the gymnasium is 118 ft across, how high is the gymnasium from the center of the court? Sol: Analyze the question and get any information you can. If the two rings are 90 ft from each other at the foci, then they are each 45 ft away from the center of the court. If the gymnasium is 118 ft across, then we would get the length of the major axis and the value of a, which is 59 ft. In semi-elliptical structures, there is no need to halve b unlike a and c. Pro tip: Find only what you need. In this case, b is being asked. No need to solve for anything else if it’s not a precursor to finding b. 50.The ceiling of a whispering gallery is in semi-elliptical shape. Two people standing at the foci can hear each other’s whisper, since sound coming from any focus bounces off the ceiling to the other focus. If the gallery is 46 ft across and 15 ft high at the center, how far apart are the two points where two people should stand to hear each other’s whisper? Sol: Since c is only the distance from the center, you need to multiply it by 2 to get the distance of the two people. 51.A room’s ceiling is in the shape of a semi-ellipse. Suppose the room is 40 ft wide and 18 ft high. How far from the end should two people stand so they can hear each other whisper? Sol: 52.A one-way tunnel has the shape of a semi-ellipse that that is 12 ft high at the center and 28 ft across. Sol: Assume that the tunnel lies at the origin. a. Will a truck that is 8 ft across and 10 ft high be able to pass through the tunnel? Sol: To see if the truck will be able to fit, find the coordinates of the upper right edge and upper left edge of the truck. Assume that the truck is placed in the origin. Substitute x of one edge into the equation to find the height of the ellipse at that particular edge. If the result is higher than y-component of the edge (in this case, 10), then the truck can fit in the tunnel. The truck is able to fit the tunnel. b. Suppose the road under the tunnel has two lanes and the truck is allowed to pass through only one lane. Will it still be able to pass the tunnel? Sol: In this case, we can use either half of the tunnel. However, for simplicity’s sake, let’s use the right one for uniformity. Instead of
57.An ellipse has foci (-6, 4) and (10, 4). If a vertex is on the line , find the equation (in standard form) of the ellipse. Sol: This ellipse is a horizontal ellipse. Therefore, the foci and the vertices will be collinear. Reading further, they would all have the same y. If y is constant at 4, find out where x is if y = 4 by substitution. 58.The figure below shows an ellipse and two circles which are tangent to each other. If the circles have equation and , find the equation in standard form of the ellipse. Sol: To get the standard form of the ellipse, you only need h, k, a, and b. Since the three figures all have the same center, then we already have h and k. The inner circle is tangent to the covertices of the ellipse. Solve its radius to get b. The outer circle is tangent to its vertices. Solve its radius to get a. 59.A graph has equation a. What is the shape of the graph
if m≠ n?
Sol: This is an ellipse. b. What is the shape of the graph
if m = n?
Sol: This is a circle. c. If a circle is considered as a special kind of ellipse, describe where the foci are. Sol: The foci meet at the center. There is only one fixed point in a circle. 60.Suppose an ellipse has foci F 1 (-c, 0) and F 2 (c, 0), where c > 0. Suppose too
that for any point P (x, y) on this ellipse, the sum of its distances from the foci is 2a, where a > c. a. Express the distances PF 1 and PF 2 in terms of c. b. Use these expressions to replace the left-hand side of the equation PF 1 + PF 2 = 2a. Let
. Derive the standard equation of the ellipse. Sol: 61.The set of all points, such that the sum of its distances from two points F 1
(− 2 √ 2 , − 2 √ 2 ¿ and F 2 ( 2 √ 2 , 2 √ 2 ¿ is 12,
is an ellipse shown in the figure below. The points F 1 and F 2 are the foci, and the center of the ellipse is the origin (the midpoint of F 1 and F 2 ). Find an equation of this ellipse. Sol: To answer this problem, we must introduce a new formula that is only used when 1) the figure is being rotated, and
- when the center is at the origin. We find out the value of a. Then, we get the distance between the foci and the center to get c. Find b using normal means. As for what angle to use, remember that a perpendicular line is always 90 degrees and the entirety of the Cartesian Plane is 360 degrees. It also moves in a clockwise manner (to the right). Thus, if the figure rotates to the left, it calls for a negative degree. 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-
