We have already seen, separately, the various types ofconics(see, for example, here, here, here, here, here). Now we try to unify them.Aconicis a curve obtained as the intersection of the surface of aconewith a plane. If I look at the conicae from the top of the cone, they all seem circular.That is why you can not distinguish the type of conic that you are observing if you do not take into account the point of view (see).a*x^2 + b*x*y + c*y^2 + d*x + e*y + f = 0, with at least one between a, b and c other than 0, in the plane x,y is aconic(see). Let Q = b^2-4*a*c; if Q = 0 is a parabola, if Q > 0 is an hyperbola, if Q <0 is an ellipse. Example:F = function(x,y)4/3*x^2+x*y/2+y^2-7*x-8*y-32PLANE(-20,20, -20,20); CURVE(F, "blue") # I change the scale ...It's anellipse. And in fact (1/2)^2-4*4/3*1 = -5.083333 < 0I can also study it using theConiccommand:conic()#Use Conic(w) where w is c(a,b,c,d,e,f) fora*= 0x^2+b*x*y+c*y^2+d*x+e*y+f#C1 and, as appropriate, xC,yC, xV,yV, xF,yF, xF1,yF1, xF2,yF2, FV, R are stored#Use fraction(xC),..,fraction(yF2) if you want to try to find fractional forms U = c(4/3,1/2,1,-7,-8,-32);Conic(U)#C1=function(x,y) 1.333333 * x^2 + 0.5 * x*y + 1 * y^2 + -7 * x + -8 * y + -32#ellipsexC = 1.967213 yC = 3.508197#dir.major ax: 298.155 ° dir.minor ax: 28.15497 °#semi-major axis 7.816129 semi-minor axis 6.005756#F1 4.327577 -0.9021841 F2 -0.3931511 7.918578 eccentr 0.6399942#Then you have the length of the string that draws the Ellipse with#StringE(xF1,yF1, xF2,yF2) [which is = major axis = distance V1-V2]#and the couple of hor/vert dist. from F1 (F2) to V1 (V2) with FVIf you want you can look for a fractional forms fraction( c(xC,yC) ) # 120/61 214/61The picture on the right:PLANE(-10,10, -5,15); CURVE(C1,"red") l2p(xF1,yF1,xF2,yF2, "red") # l2p is a thin line between the 2 points (see) text(1.5,2,"C"); text(-1.5,7,"F1"); text(3,-2,"F2") POINT(xF1,yF1,"blue"); POINT(xF2,yF2,"magenta"); POINT(xC,yC,"seagreen") POINT(xF2-FV[1],yF2+FV[2],"black"); POINT(xF1+FV[1],yF1-FV[2],"brown")# To have the sum of distances focus1-curve-focus2: StringE(xF1,yF1,xF2,yF2) # 15.63226 or 7.816129*2 PLANE(-10,10, -5,15); CURVE(C1,"blue"); POINT(xF1,yF1,"red"); POINT(xF2,yF2,"red") # I trace one of the point with y=10: K1 = function(x) F(x,10); solution(K1,0, 3,5) # 3.842333 POINT(solution(K1,0, 3,5),10, "black") polyl(c(xF1,solution(K1,0, 3,5),xF2), c(yF1,10,yF2), "black") # Now, I can trace the ellipse even in this way: K = function(x,y) point_point(xF1,yF1,x,y)+point_point(xF2,yF2,x,y)-StringE(xF1,yF1,xF2,yF2) CUR(K,"red")# If you want to see how to study theColosseumsee here.---------- ---------- ---------- ---------- ---------- ---------- ---------- ----------Let's look at another example:4/3*x^2-y^2+x*y/2-7*x-8*y-32= 0conic()#Use Conic(w) where w is c(a,b,c,d,e,f) fora*= 0x^2+b*x*y+c*y^2+d*x+e*y+f#C1 and, as appropriate, xC,yC, xV,yV, xF,yF, xF1,yF1, xF2,yF2, FV are stored#Use fraction(xC),..,fraction(yF2) if you want to try to find fractional formsV = c(4/3,1/2,-1,-7,-8,-32);Conic(V)#C1=function(x,y) 1.333333 * x^2 + 0.5 * x*y + -1 * y^2 + -7 * x + -8 * y + -32#hyperbolaxC = 3.223881 yC = -3.19403#dir.trasv.ax 2.359818 0.25 dir.coniug.ax -0.25 2.359818#V1 7.934077 -2.69503 V2 -1.486316 -3.69303#F1 10.40555 -2.433201 F2 -3.957791 -3.954859 eccentr 1.524707#In C2 there is the curve formed by the 2 asymptotes fraction( c(xC,yC) ) # 216/67 -214/67 PLANE(-10,15, -15,10); CURVE(C1,"blue") POINT(xC,yC,"seagreen"); CUR(C2,"red"); POINT(xC,yC,"seagreen") POINT(xF1,yF1,"red"); POINT(xF2,yF2,"red") l2p(xF1,yF1,xF2,yF2, "seagreen")text(3,-1,"C",cex=0.8); text(-3.4,-2.5,"F2",cex=0.8); text(12,-3.8,"F2",cex=0.8)It's ahyperbola. And in fact (1/2)^2+4*4/3*1 > 0All the hyperbolae whose asymptotes are represented by the equation C2(x,y)=0 can be represented by an equation of the form C2(x,y)=k. Fixed a point we can find the hyperbola passing trough it. In this case I can find the hyperbola passin trough V1:CC2 = function(x,y) C2(x,y)-Q E = function(x) C2(xV1,yV1)-x; Q = solution(E,0, -100,100) CUR(CC2, "cyan") # I see (above) that the curve is the same---------- ---------- ---------- ---------- ---------- ---------- ---------- ----------Let's look at another example:x^2+y^2-2*x*y+4*x+4= 0conic()#Use Conic(w) where w is c(a,b,c,d,e,f) fora*= 0x^2+b*x*y+c*y^2+d*x+e*y+f#C1 and, as appropriate, xC,yC, xV,yV, xF,yF, xF1,yF1, xF2,yF2, FV are stored#Use fraction(xC),..,fraction(yF2) if you want to try to find fractional formsW = c(1,-2,1,4,0,4);Conic(W)#C1=function(x,y) 1 * x^2 + -2 * x*y + 1 * y^2 + 4 * x + 0 * y + 4#parabolaxV = -1.25 yV = -0.25 dir.ax.symm = -1 -1#F -1.5 -0.5 In C2(x,y) there is the directrix PLANE(-4,1/2, -3,1.5); CURVE(C1, "blue") CUR(C2, "red"); POINT(xV,yV, "brown"); POINT(xF,yF, "red") l2p(xF,yF,xV,yV, "seagreen")

eccentricity = 1 |

Can you foresee that the following commands provide the figure on the left? How do you do? G1 = function(x,y) x-y-1; G2 = function(x,y) x+y-1 S = c( 1,-1, 1, 1, 1, 1 ) eqSystem(S) # 1 0 BF=2.5; HF=2.5; PLANE(-1.5,3.5, -2.5,2.5) CURV(G1,"blue"); CURV(G2,"blue"); POINT(1,0, "brown") G = function(x,y) G1(x,y)^2+3*G2(x,y)^2-4 H = function(x,y) abs(G1(x,y))+3*abs(G2(x,y))-4 CURVE(H, "red"); CURVE(G, "seagreen") |

Can you trace the figure on the right, if you know that the equation of the curve is x^2-2*x*y=4*y? [ use the command Conic( c(...) ) ] |

Can you trace the figure on the left, if you know that the equation of the curve is x^2+2*x*y+y^2+4*x=5? [ use the command Conic( c(...) ) ] |

Trace asymptotes and foci of the equilateral hyperbola 2*x^2+3*x*y-2*y^2-7*x+y = 0 [ use the command Conic( c(...) ) ] |

Trace the parabola with focus (0.1, 1.2) and directrix that passes through points (-1/2,0) and (0,-1/4) [ Use distance point_point and point_line Compare with 4*x^2-4*x*y+y^2-2*x-14*y+7 = 0 ] |