Graphing Conic Sections

Transcription

1 Grphing Conic Sections Definition of Circle Set of ll points in plne tht re n equl distnce, clled the rdius, from fixed point in tht plne, clled the center. Grphing Circle (x h) 2 + (y k) 2 = r 2 where (h, k) is the center nd r is the rdius. Grph the circle descried y the eqution (x 4) 2 + (y + 2) 2 = 4. Answer: Strt y finding the center of the circle (h, k). x h x 4 h = 4 y k y + 2 y (-2) k = -2 Thus, the center is locted t (4, -2). Since r 2 = 4, tking the squre root of oth sides yields r = 2. Therefore, the grph looks like the following grph: Definition of Prol The set of ll points in plne equidistnt from given point, clled the focus, nd given line, clled the directrix. The prol either hs verticl xis (s shown) or horizontl xis (prol opens sidewys). xis focus p vertex directrix

2 Grphing Prol Horizontl Axis x h = 1 y k ( ) 2 (h, k) vertex of prol p directionl distnce from vertex to focus Verticl Axis y k = 1 x h ( ) 2 Write the eqution nd grph the prol with focus (-1, 2) nd directrix y = 6. Answer: To write the eqution, you need the coordintes of the vertex of the prol (h, k) nd the vlue of p, the directed distnce from the vertex to the focus. The vertex lies on the xis hlf wy etween the focus nd directrix. Since the directrix hs the eqution y = 6, it is horizontl line, which mens the prol hs verticl xis. So, the x-coordinte of the vertex is the sme s the x- coordinte of the focus. The y-coordinte of the vertex is the verge of the y-coordinte of the focus nd the y-coordinte of the directrix. So, the y-coordinte of the vertex is 4 (2 + 6 = 8 2). Thus, the vertex hs coordintes (-1, 4). The vlue of p is the directed distnce from the vertex to the focus. Since the vertex is t (-1, 4) nd the focus is t (-1, 2), the vlue of p is -2. Plugging in the vlues h = -1, k = 4, nd p = 2 into the vertex-form of the eqution for the verticl xis yields y k = 1 ( x h) 2 y 4 = 1 ( x + 1) 2 y 4 = 1 ( x + 1) 2 4( 2) 8 The grph of the prol is s follows:

3 Definition of n Ellipse The set of ll points in plne in which the sum of the distnces from two fixed points, clled the foci, is constnt. Co- Mjor Axis Center Minor Axis Co- Grphing n Ellipse Horizontl Mjor Axis Verticl Mjor Axis + = 1 where 2 2 = c 2 (h, k) center of ellipse distnce from center to vertex distnce from center to co-vertex c distnce from center to focus + = 1 ( x + 2) ( y + 1) Grph the ellipse given y the eqution + = Answer: Since the eqution is in stndrd form, you should e le to determine the center of the ellipse, the vlues for nd, nd wht direction the mjor xis points. Since 2 > 2, 2 = 25 nd 2 = 16. Thus, = 5, nd = 4. These vlues help shpe the grph. Since 25 is under the y-vrile, the mjor xis is verticl. Solve for h nd k to find the coordintes of the center. x + 2 x (-2) x h h = -2 y + 1 y (-1) y k k = -1 Thus, the center is t (-2, -1), the distnce from the center to the vertex is 5, nd the distnce from the center to the co-vertex is 4.

4 Definition of Hyperol The set of ll points in plne in which the difference etween the distnces from two fixed points, clled the foci, is constnt. Asymptote Trnsverse Axis Center Conjugte Axis Asymptote Grphing Hyperol Horizontl Trnsverse Axis Verticl Trnsverse Axis where = c 2 (h, k) center of hyperol 2 length of trnsverse xis 2 length of conjugte xis c distnce from center to focus ( y k) ( x h) Find the vertices of the hyperol given y the eqution 4y 2 9x 2 54x + 8y = 41. Then, sketch the grph. Answer: First, write the eqution is stndrd form y completing the squre for oth the x nd y-vriles. Rewrite the eqution to put vriles together: 4y 2 + 8y + 9x 2 54x + = Fctor out coefficients of squred vriles (*wtch for negtive numers): 4(y 2 + 2y + ) 9(x 2 + 6x + ) = Divide the middle numer y 2 to complete the squre: 4(y + 1) 2 9(x + 3) 2 =

5 Rewrite previous eqution with correct vlues: 4(y 2 + 2y + _1_) 9(x 2 + 6x + _9_) = Use the distriutive property to get correct vlues: 4y 2 + 8y + _4_ 9x 2 54x + _-81_ = 41 + _4_ + Simplify nd write in fctored form: 4(y + 1) 2 9(x + 3) 2 = -36 Divide oth sides y -36 to get eqution in stndrd form: ( y ) ( x ) = ( x + 3 ) ( y + 1 ) 4 9 ( y + 1) ( x + 3) _-81_ + = Since the eqution is now in stndrd form, you cn find the center nd the vlues for,, nd c in order to help grph the eqution. Since the x-vrile is with the positive frction, the hyperol horizontl trnsverse xis. Also, 2 = 4, since 2 is lwys the denomintor of the positive frction. Since 2 = 4 nd 2 = 9, tht mens = 2 nd = 3. To find the center, solve for h nd k. x + 3 x (-3) x h h = -3 y + 1 y (-1) y k k = -1 The vertices re units to the left nd right of the center (since there is horizontl trnsverse xis). Thus, the vertices re locted t (-3 ± 2, -1), or (-3 + 2, -1) nd (-3 2, -1), which re the points (-1, -1) nd (-5, -1). You cn use the vlues of nd to help form the ox to drw in the symptotes. Since = 2 nd = 3, the ox will hve width of 4 (2 = 2 2) nd height of 6 (2 = 2 3). Thus, the grph should look like the following: