Conic-answers

Solutions to the Sample Problems from Unit 10: Conics
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 * 1) We know that the room is shaped like an ellipse with a major axis of 68 feet and a minor axis of 32 feet. The question asks for the distance between the foci. Given that in an ellipse, the major radius and minor radius form a Pythagorean triple with the focal length (b^2 + f^2 = a^2), we can set up an equation to find f, the focal length. a and b are half of the major and minor axes respectively so 16^2 + f^2 = 34^2. If we solve for f, we get f = 30 or -30. A length can't be negative so we can eliminate -30. Finally, the focal length will be half the distance between the foci so we can double f to get a final answer of 60 feet.
 * 2) __Foci of ellipse: (-sqrt(5),0), (sqrt(5),0) Foci of hyperbola: (sqrt(13),0), (-sqrt(13),0) The solution to the system is (1.52,0.68), (-1.52, -0.68), (1.52, -0.68), and (1.52, -0.68).__ Explanation: The foci of an ellipse can be found using a and b and the pythagorean theorem. We then plug in the values 3 and 2 for a and b respectively and get the equation 2^2 + f^2 = 3^2 and f=sqrt(5) or -sqrt(5). For the foci of the hyperbola, we use the equation f^2=a^2+b^2 and plug in 2 and 3 for a and b respectively and get the eqation 2^2 + 3^2 = f^2 and f=sqrt(13) or -sqrt(13). In order to solve the system, we have to find the equation for both the ellipse and the hyperbola. The parent function of the ellipse is x^2 + y^2 = 1. The only transformation applied is a stretch of 3 to the horizontal and 2 to vertical, so this changes the parent function to (x^2)/3 + (y^2)/2 = 1. For the hyperbola, using this same train of thought, the parent function x^2 - y^2 = 1 changes to (x^2)/2 - (y^2)/3 = 1. Solving for this system of equations gives us 1.52 and -1.52 as x-coordinates and also gives us 0.68 and -0.68 as y-coordinates.
 * 3) The easiest way to solve this is to put the two original kids on the x-axis at the points (5,0) and (-5,0). Thanks to the knowledge about the length of the string (7+7=14, the total string is not 7 feet long), we know that 2 other points are at (7,0) and (-7,0) with the two original friends being the foci of the ellipse. Since we also know that the y-stretch, the x-stretch, and the distance from the center to the focus form the pythagorean theorem, we know that 5^2+B^2=7^2. (with B being the y-stretch). This ends with B=24^1/2 so that is the y-stretch. This makes the equation of the ellipse (x/7)^2 + (y/24^1/2)^2 = 1. If the + was turned into a -, then a hyperbola would be formed with the distance between the 2 friends being the constant.
 * 4) You know that the new center of the circle must be (5, -2) because of the translation. Looking at the x and y stretches (the 4 and 3) you can find four points on the edge of the circle. These four points are (9, -2), (1, -2), (5, 1), and (5, -5). The farthest apart any two points on the curve could be is 8 units. You know this quickly because that is the length of the major axis.
 * 5) (x+5)^2 = 22-y^2
 * 6) ( (x-3)^2 /20) - ( (y-1)^2 /4) =1 AND ( (x+7) ^2 /16) - ( (y+3)^2 /20) since you know the foci of an ellipse are on the major axis, you can either have a vertical ellipse or horizontal ellipse. once you find the other vertex you can find the center of both possible ellipses and from there use pythagorean theorem to get the other stretch. Then all you have to do is transform it!
 * 7) The values of the y^2 coefficient and the x^2 coefficient are the same sign, non-zero, and different numbers, so it is an ellipse. Shiny? (<-- lol -Alexis)
 * 8) 7.8 is the answer. We know this because the distance between the vertices of the major axis is always equal to the sum of the distances from a point on the ellipse to the two foci. Therefore, since the distance between the vertices is 12, and we know the point is 4.2 away from a focus, the distance to the other focus plus 4.2 must equal 12. 12 - 4.2 = 7.8.
 * 9) When the average is found of the points ((5,4) and (-1,4)) the center of the circle can be found, and it ends up being (2,4). Then either using the distance formula or pythagorean theorem the radius can be found, it ends up being 2. Therefore the equation of the circle is (x-2)^2+(y-4)^2=(3)^2
 * 10) ((y-2)/1)^2+((x-1)/4)^2=1
 * 11) The y intercepts are (0,9) and (0,-9). The x intercepts are (3,0) and (-3,0). The foci are (0,(72)^1/2) and (0,-(72)^1/2). The y axis is the major axis.
 * 12) (1/4)(y-4)^2 +3 = x
 * 13) This is an ellipse. The major axis is on the y axis and the minor axis is on the x-axis. This graph has no translations, so its center remains at (0,0). You stretch the x points by 4 and the y-coordinates by 6.
 * 14) It is a parabola because there is not a y^2 coefficient. It is vertical because the equation starts with y= not x=.
 * 15) From slide of given ellipse, you can deduce 1=((x-2)/a)^2+((y-2)/b)^2. Given that b^2=focal length^2+a^2 (as it is a vertical ellipse), you can plug in any value for a or be and solve for the the other. The focal length is the distance from center to vertex in the given parabola (5).