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1 MA00 Clculus nd Bsic Liner Alger I Chpter Coordinte Geometr nd Conic Sections Review In the rectngulr/crtesin coordintes sstem, we descrie the loction of points using coordintes. P (, ) P(, ) O The distnce d etween two points P nd P is given d,, The midpoint M of line segment etween P nd, The slope m of line joining P nd, P is given,. P, is given,. m. Note: m > 0 for stright lines which slope upwrd to the right nd vice vers. m is undefined for verticl lines nd m = 0 for horizontl lines. Let L nd L e two lines with slopes m nd m. Then L // L if nd onl if m m ; L if nd onl if m m. L Specil forms of the eqution of stright line: ) Generl form of stright line: A + B + C = 0, where A, B, C re constnts such tht A nd B cnnot oth equl to zero. ) Two points form: The eqution of the stright line pssing through the points P nd = ) Pointslope form:. P (, ) The eqution of the line with slope m pssing through point P m. 4) Slopeintercept form: The eqution of stright line cn lso e written in the form nd the intercept of the stright line respectivel.,, is Q, is m c, where m nd c re the slope
2 Emple Find the eqution of stright line which stisfies ech of the following conditions: () joining P(, 4) nd Q(, ); () perpendiculr to L : + 5 = 0 nd cuts the is t (5, 0). () The eqution of stright line L is given () Eqution of L in slopeintercept form is = +. Slope of L is m =. Slope of the required line L is m =, since m m =. The eqution of stright line L is given 0 = Eercise: ) Find the coordintes of the foot of the perpendiculr from P(6, ) to the stright line joining Q(, ) nd R(, 4). ) Find the shortest distnce from the origin O(0, 0) to the line joining A(, 4) to the point of intersection of the line + 6 = 0 with the is. Review on Polr Coordintes The polr coordintes of point P is represented s r,, where r is the distnce of the point from the pole nd is n ngle formed the polr is nd r from the pole through the point. P(r, ) vectoril ngle Polr is O origin (pole) (initil is)
3 The reltions etween Polr nd Rectngulr Coordintes re: r cos r sin r tn Question: Convert ech of the following points from polr coordintes to rectngulr coordintes., 45, 0, 60, 40 (i) (ii) 7 (iii) 6 (iv) Emple Determine of the polr coordintes of the following points with the given rectngulr coordintes. Epress in the rnge () P, () P, (c) P,. () Notice tht, P lies on the first qudrnt. P In OP Q, PQ tn Second qudrnt First qudrnt OQ 60 O Q 60 P, lies on the second qudrnt. In OP Q, () Notice tht P (c) Notice tht, Q P Question: Convert 5, 5 P Q tn OQ P lies on the third qudrnt. In OPQ, PQ Q O O tn OQ 45 Third qudrnt P from rectngulr to polr coordintes with Fourth qudrnt
4 Conic Sections Mthemtics is present in different spects, for emple, the movements of plnets, ridge nd tunnel construction, mnufcture of lenses for telescopes nd so on. The mthemtics ehind these pplictions involves conic sections. Conic sections re curves tht result from intersecting right circulr cone with plne. The following figure illustrtes the four conic sections: the prol, the circle, the ellipse nd the hperol. Prol Circle Ellipse Hperol Question: Wht else could e otined ecept the four figures shown ove when plne intersects right circulr cone? A. Prole Definition: A prol is the set of ll points in plne tht re equidistnt from fied line, the directri, nd fied point, the focus. The equtions of prol with focus t the point F, 4. 0 nd the directri, where 0, is 4
5 Proof: The point P(,) is on the prol if nd onl if it is equidistnt from the directri F 0,, i.e. focus 0 4 nd the The prol 4 0 ) hs the following properties: (i) It lies ove the is. (ii) It is smmetricl out the is, which is clled the is of smmetr of the prol. (iii) It cuts the is t the origin O, which is clled the verte. (iv) As the vlue of increses, the prol opens wider. Depending on the loction of focus nd the orienttion of the directri, there re other forms of prole (with the verte t the origin nd > 0):
6 Emple The verte nd the is of smmetr of prol re the origin nd the is respectivel. If the prol psses through the point (6, ), find its eqution. Since the prol is smmetricl out the is, we let its eqution e This prol psses through the point (6, ), 46, get. Hence the eqution of the prol is 8 4, i.e Trnsltions of Prole The grph of prol cn hve its verte t h, k rther thn t the origin. Horizontl nd verticl trnsltions re ccomplished replcing with h nd with k in the stndrd form of prol s eqution. Eqution Verte Ais of Smmetr Description k 4 p h h, k k If p 0, opens to the right; if p 0, opens to the left. h 4 p k h, k h If p 0, opens upwrd; if p 0, opens downwrd. More on c (i) The grph of the qudrtic eqution c is prol which opens upwrd when 0 nd opens downwrd when 0. (ii) The prol  intersects the is t distinct points iff the discriminnt, 4c 0 ;  touches the is t point iff 4c 0 ;  does not cut the is iff 4c 0. 6
7 B. Circle Definition: A circle is the set of ll points in plne tht the distnce of the point from fied point is constnt. The fied point is clled the centre nd fied distnce is clled the rdius of the circle. The equtions of circle with centre t the origin O 0, 0 nd rdius r is r. This is known s the stndrd form of the eqution of circle centered t the origin. Proof: Leve s eercise. The equtions of circle with centre t the point C h, k nd rdius r is h k r.. Proof: Leve s eercise. Emple 4 Find the equtions of the circles with the following centres nd rdii. () Centre t, 0,, rdius: units 0, rdius: 7 units () Centre t () The eqution of the circle is: i.e. 49 () The eqution of the circle is: i.e. Emple 5 Find the centre nd rdius of the circle represented ech of the following equtions. () 5 0 () () Rewrite the eqution in stndrd form: Centre is t, 5, rdius is 0 units
8 () Rewrite the eqution in stndrd form: Centre is t 4, , rdius is 7 units. Remrk: The technique used in () is clled completing the squre. Emple 6 Find the eqution of the circle centred t the point 4, nd pssing through the origin. The rdius of the required circle is Thus, the eqution of the circle is Emple 7 Find the eqution of the circle through P(, ), Q(, 4), R(7, 6). Method I Since the side P Q hs slope m = 4 = nd its midpoint is (, ). The side P R hs slope m = 6 7 = nd its midpoint is (4, 4). The perpendiculr isectors of P Q nd P R re =, or = 5 nd 4 4, or
9 Solving the equtions () nd (), we otin 5 0 5, 0, 5. These two perpendiculr isectors meet t 0, 5, which is the centre of the circle. The rdius of the circle is given units. The eqution of the circle is 0 5 0, Which reduces to Method II Let 0 e the eqution of the circle through the points P(, ), Q(, 4), R(7, 6). Then : 0 : : 0, 0 From 4, we hve 0 From, we hve The eqution of the circle is
10 C. Ellipses Definition: An ellipse is the set of ll points P in plne tht the sum of whose distnces from two fied points, F nd F, is constnt. The two fied points re clled the foci (plurl of focus). The midpoint of the segment connecting the foci is the centre of the ellipse. The equtions of n ellipse with foci t the points F c, 0 nd F c, 0, where c 0, is, where 0, c nd is the sum of the distnces from n point on the ellipse to the two foci. The eqution ove is the stndrd form of the eqution of n ellipse centered t the origin. Proof: The point P(, ) is on the ellipse if nd onl if the sum of distnces from F 0 nd 0 i.e. c + c Simplifing, get c c Oserved tht F F c PF PF, so tht c nd c 0. For convenience, let c. We hve c, F c, is, The ellipse ( 0) hs the following properties: (i) It is smmetricl out the is, the is nd the origin. (ii) Its foci re F c, 0 nd F c, 0, nd cn e found the eqution c. (iii) It cuts the is t A, 0 nd A, 0, nd the is t B 0, nd B 0,. These four points re clled the vertices of the ellipse. The line segments A A nd B B intersect t the origin O, which is the centre of the ellipse. The length of the line segment A A is greter thn tht of the line segment B B. 0
11 Depending on the loction of foci, there is nother form of ellipse (with the centre lso t the origin): where 0,, c. Emple 8 Find the coordintes of the vertices nd the foci of ech ellipse, nd sketch its grph. () () 8 8 () Rewrite the eqution of the ellipse s, We hve nd. Note tht the eqution represents n ellipse with the centre t the origin. The vertices of the ellipse re, 0,, 0, 0, nd 0,. Moreover, c 5, i.e. c 5 (c is positive.) The foci of this ellipse re F 5, 0 nd F 5, 0. 0 The sketch of the ellipse is shown on right = 6 () Rewrite the eqution of the ellipse s, so tht nd. Note tht the eqution represents n ellipse with the centre t the origin. The vertices of the ellipse re, 0, 0, 0,.,, 0 nd Moreover, c 7, i.e. c 7 (c is positive.) The foci of this ellipse re 0, 7 nd 0, 7. The sketch of the ellipse is shown on right.
12 Emple 9 Find the eqution of the ellipse whose centre is the origin nd the ellipse psses through the points, 0,. nd Let the eqution of the ellipse e. p q Since the ellipse psses through the point As it lso psses through the point,, we hve Hence the eqution of the ellipse is 8 6., 0 0, we hve, get p q 8, i.e. q 6. q p. If the ellipse is trnslted h units to the right nd k units upwrd so tht its new centre is t C h, k, then the equtions of the new ellipse ecomes h k.
13 D. Hperole Definition: A hperol is the set of ll points in plne tht the difference of whose distnces from two fied points, clled the foci, is constnt. The eqution of hperol with foci t the points F 0 nd 0 c, F c,, where c 0, is, where, 0, c nd is the solute difference of the distnces from n point on the hperol to the two foci. The eqution ove is clled the stndrd form of the eqution of hperol centered t the origin. Proof: The point P(, ) is on the hperol if nd onl if the solute difference of whose distnces from F c, 0 nd F c, 0 is, c c i.e. Simplifing, get c c Oserved tht F F c, so tht c 0. For convenience, let c. We hve
14 The hperol (, 0 ) hs the following properties: (i) It is smmetricl out the is, the is nd the origin. (ii) Its foci re F c, 0 nd F c, 0, nd cn e found the eqution c. The midpoint of these two foci is the origin O, which is the centre of the hperol. (iii) It cuts the is t A, 0 nd A, 0, which re clled the vertices of the hperol. (iv) As nd get lrger, the two rnches of the grph pproch pir of intersecting stright lines nd. These re clled the smptotes of the hperol. Depending on the loction of foci, there is nother form of hperol (with the centre lso t the origin):, 0, c, Emple 0 Arrnge 9 4 = 44 into the stndrd form of hperol. 9 4 = If the hperol is trnslted h units to the right nd k units upwrd so tht its new centre is C h, k, then the equtions of the new hperol ecomes t h k. 4
15 When the eqution of conic section is epressed in the form A C D E F 0, one cn identif the conic section completing the squres so tht the eqution is epressed into the corresponding stndrd form. Emple Clssif the tpe of conic section descried ech of the following equtions using completing the squres. () () () The eqution represents n ellipse. () The eqution represents circle. 4 Prmetric Equtions of Conic Sections f t Consider the pir of equtions. g t To ech rel vlue of t, there corresponds to point (, ) in the Crtesin plne. The point (, ) moves nd trces curve s t vries. f t, g t The pir of equtions prmeter. f t g t curve C is clled prmetric equtions nd the independent vrile t is clled 5
16 The following tle lists the equtions of conic sections in rectngulr coordinte form nd in prmetric form. Tpe of Conics Prol Tpe Prol Tpe Circle, centred t (0,0) nd rdius r Eqution in Rectngulr Coordinte Form 4 (or 4 ) 4 r Ellipse Ft ;( ) Ellipse Thin ;( ) Hperol (EstWest Openings) Hperol (NorthSouth Openings) Rectngulr Hperol c ; c is constnt Eqution in Prmetric Form t, t t t, t t r cos t, 0 t r sin t cost ;( ), sin t 0 t cost ;( ), sin t 0 t sec t, t tn t sec t, t tn t t c, t, t 0 t 6
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