Parabola - Algebra Lecture Note

classmateDatePageParabolaItisthelocusofadistancefrommovingpointwhoseaisfixedequalpoint(focus)tothefixeddistancefromaune (directrix)StandardEquation(h-a)2 + k2 = (h-(-a)/P(h,k)doubleu(h --a)2 + k2 = (h+a)2ordinateaxisk2 = 4ah = y 2 = 4axEs(a,o)vertexchord(axis passes thru S, + to dirx)tartice(vertex = intx axis curve) (x=-a)rectum focal chord(doubledouble ord &ordinateischord1toaxis)focal chordFormsi2 = yaxy2 = -4axx2 = 4ayx-a-may--IS(shape)sisS(vertex)(0,0)(0,0)(0,0)(0,0)(focus)(a,o)(-a,o)(0,a)(0,-a)(dirx)x = -ax = ay=-ay=a(axis)y=0y=0x=0x=0(transite)x=0x = 0y=04=0( end of pts)(a, z2a)(-a, + 2a)(=2a,a)(=2a,-a)(venatth of4a4a4a4a(parameto)(at2,2at)(-at2, 2at)(2at,at)(2at,-ct"( focal dist)at'+aa + at2at' taa+atz(Ih - cal)(a -catz): Pagenote length of LR = 2 (+ dist of S from dineIf y' = kx, length of LR = k.Two parabolas are equal it thelength of LR is equalalso ifthesegmented intercepted bythe parabola y2 = 4ax withtheline Lx + my + n = D subtendsa right angle atthe vertex,theline always passes three (4a DproofHomogenisingtheparabolaequation we gety2 = 4ax &Lx + my=1- ny2 = 4ax ( lx +my-n)(4al) x2 + (4am) xy + ny2 =:0Since the two linis are perp(4al+n=0x l +ym + n = 0is 4a=0Thus, theline passes then (4a , 0).AnotherdefinitionofparebolaisRrIt is the locus of theRcentre of a circle thattouches a given line & para botagiven circle externallyt(direz( give line & circle don't inbx) Shifted ParabolaIf the vertex is A (u, k) and the axisisparallel to y = 0,.(y-k)2 = 4a(x. -h)1A42 = 4a X <(h,k)(use shifting of origin)E(vertex) II (n, k)focus III (a th, k)tav = x = hdirx III x = h - aextremitiesof LR III (a +h, I 2a +k)length LR of in4aparametricwords II (at2+h, 2at +k)focal dist in a + at2Quadratic Equationformx = Ay- + By + Cy = Ax2 + Bx+ Clength of LR1/1411/1A1axis(11 to) x axis(II to) y axisvertex(-P/UA, -8/2A)(-B/2A , -D/MADA70rightward openupward openA <0leftward opendownward open DatePageGeneral EquationLet S(x, B) be the focus &lx + my + n = 0 be the directrixLet the locus be p(h, k). This :V(h-a)2 + (le - B)2 = I lh + mk. + ull2 +thi2Squaring both sides(l2 + m2) (1x-x)2 + (y-B)2) = (lx + my+n)2Upon expansion & simplification,m2x2 + lzyz - 21mxy +(mx-ly)2 + 2gx + 2ty + c = 0 DatePageParabolaWHTPerpendicularLinesThetwopeeplinesh iPserveastheaxisN&axistauoftheparabolaM(PM)2 = 4a(PN)(dist from axis) = 4a (dist from tau)tanlength of LRPositionofaPointwrtParabolaor/Let S(x, y) = y2 - 4ax and Six, ys = x - 4ayIfPlies :insidetheparabolaSp<0outsidetheparobolaSp > OontheparabolaSp = 0ChordsinaParabolaTheequationofachordjoiningthepointsP(t,)&Q(t2)is:(ts+t2)=2x+y2at t2(parametricproof)ThefocalchordinachordthatpasiesthruthefocusS(a,o)iny2=4ax.IfP(t)&Q(t2)aretheendpointsofafocalchordt,t2-1.(put a,o in aboveequation OR mecollincarity conda) PageThe extremeties of the focal chordare(at22at)&(a(z)22a())(:titz=.Thelengthofthefocalchordis:IPO =PS+soN/p(t)22= PN + QMtan ==ti+tzt-t=a + at2+a+ast2MQ(H)= a (t + 1/4)2= Lacose c20min = 4aThesmallestfocalchordistheLRThesemi-LRistheharmanicmeanofPS&QSi.e.2/(2a)=1/PS+,1/asAcircledescribedonthefocallengthasadiametertouchesthetav.Acircledescribedona'focalchordasadiametertouchesthedirx. DatePageIntersectionofLine&Parabolaletthelinebey=mx+candthe2parabolabey 2 = 4ax Solving simultaneously we get :(mx+c)e = 4axmx2 + 2cum - -2a)x + c2=0(max two urtx polutes)D=6 a (a - cm)If :Dc0(a=cm)thenthecondition,oftangency is satisfiedone intoD>0 then there are two intuD