Some long time ago, I worked out for fun the reverse problem: given a general quadratic Axx + Bxy + Cyy + Dx + Ey + F, find foci, directrices, and eccentricity. It turned out to be somewhat interesting... 30 Aug 1982 focus u,v directrix px + qy + r = 0 eccentricity g (x-u)^2 + (y-v)^2 = g^2 (px+qy+r)^2 / (p^2+q^2) = (Px+Qy+R)^2 where P := gp(p^2+q^2)^-.5 Q := gq(p^2+q^2)^-.5 R := gr(p^2+q^2)^-.5 x^2-2ux+u^2+y^2-2vy+v^2 = P^2 x^2 + 2PQxy + Q^2 y^2 + 2PRx + 2QRy + R^2 x^2(1-P^2) + xy(-2PQ) + y^2(1-Q^2) + (-2u-2PR)x + (-2v-2QR)y + (u^2+v^2-R^2) = 0 a b c d e f A/k B/k C/k D/k E/k F/k a+c = 2-(P^2+Q^2) = 2-g^2 Note: g = 0 <=> P=Q=0 b^2-4ac = 4 P^2 Q^2 - 4 - 4 P^2 Q^2 + 4 P^2 + 4 Q^2 = 4g^2 - 4 Case 0: B^2-4AC = 0 [Parabola] g^2 = 1 a+c = 1 A+C = k k = A+C [but if 0, then A=B=C=0, so Dx+Ey+F = 0, giving line unless D=E=0, in which case plane if F = 0, else nothing] <a,b,c,d,e,f> = <A,B,C,D,E,F>/k Case 1: B^2-4AC <> 0 S := (A+C)^2/(B^2-4AC) + 1 = (B^2+(A-C)^2)/(B^2-4AC) = (b^2+(a-c)^2)/(b^2-4ac) = g^4 / (4g^2-4) g^2 = 2S +- 2(S^2-S)^.5 = 2(S +- (S(S-1))^.5) 2-g^2 = -2(S-1 +- (S(S-1))^.5) with following choice of sign: B^2-4AC < 0 [g^2 < 1] [Ellipse] S < 0 Since g^2 >= 0, must use + B^2-4AC > 0 [g^2 > 1] [Hyperbola] S > 1 Now, if A+C = 0, S(S-1)=0 so sign is irrelevant. Otherwise (Case a below), sgn(k) = sgn(A+C) sgn(2-g^2), while (from Case 1 below) sgn(k) = sgn((B^2-4AC)F+AE^2+CD^2-BDE), so sgn(2-g^2) = sgn(A+C) sgn((B^2-4AC)F+AE^2+CD^2-BDE). So, must use same sign as -(A+C)[(B^2-4AC)F+AE^2+CD^2-BDE] Case a: A+C <> 0 k = (A+C)/(a+c) = (A+C)/(2-g^2) <a,b,c,d,e,f> = <A,B,C,D,E,F>/k Case b: A+C = 0 0 = a+c = 2-g^2, so g^2 = 2 and b^2-4ac = 4g^2-4 = 4 k^2 = (B^2-4AC)/(b^2-4ac) = (B^2-4AC)/4 k = +-(B^2-4AC)^.5 / 2 By Case 1 below, must use same sign as [(B^2-4AC)F+AE^2+CD^2-BDE] <a,b,c,d,e,f> = <A,B,C,D,E,F>/k P = (1-a)^.5 Q = -b/(2P) = -b/(2(1-a)^.5) if P <> 0 = (1-c)^.5 if P = 0 u = -d/2-PR v = -e/2-QR f = u^2 + v^2 - R^2, while P^2+Q^2 = g^2, so (g^2-1)R^2 + (dP+eQ)R + (d^2+e^2)/4 - f = 0 Case 0: R = (f - (d^2+e^2)/4) / (dP+eQ) Case 1: R = (-(dP+eQ)+-((dP+eQ)^2-(g^2-1)(d^2+e^2-4f))^.5)/(2(g^2-1)) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Note: Radicand = ae^2 + cd^2 + (b^2-4ac)f - bde Note: | a b/2 d/2 | det := | b/2 c e/2 | = -(ae^2 + cd^2 + (b^2-4ac)f - bde)/4 | d/2 e/2 f | This determinant is invariant under skew, rotation, reflection, and translation, as can be seen using homogeneous coordinates: [ a b/2 d/2 ] [x] [x y 1] | b/2 c e/2 | |y| = ax^2 + bxy + cy^2 + dx + ey + f [ d/2 e/2 f ] [1] Funny Cases: Ellipse: det = 0: single point, given by "Center" det > 0: empty Parabola: det = 0: A=B=C=0: D=E=0: F=0: plane F<>0: empty else: single line (Dx+Ey+F = 0) else [k = A+C] (mx+ny+o)^2+p, where m = a^.5, n = c^.5, if a<>0, o = d/2m else c<>0 and o = e/2n p = f-o^2 [f-d^2/4a or f-e^2/4c] p>0: empty p=0: double single line (mx+ny+o = 0) p<0: parallel lines (mx+ny+o+-(-p)^.5) Hyperbola: det = 0: pair of crossing lines, given by "Asymptotes" Finding the Center: Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 d/dx = 2Ax + By + D = 0 d/dy = Bx + 2Cy + E = 0 *******Solvable only in Case 1: [2A B] [s] = [D] [s] = ____1____ [2C -B] [D] = ____1____ [2CD-BE] [B 2C] [t] [E] [t] B^2 - 4AC [-B 2A] [E] B^2 - 4AC [2AE-BD] Asymptotes of Hyperbola: (A(x-s)+(B+-((B^2-4AC)^.5)/2)(y-t))(((B+-(B^2-4AC)^.5)/2)(x-s)+C(y-t)) where we choose signs to be + if B >= 0, - otherwise, to ensure nonzeroness. On 08-Nov-18 08:43, Cris Moore wrote:
I’m going to talk to my daughter’s 8th grade class about conic sections. I want to focus on foci (ha), and how curves with beautiful geometric descriptions also have nice algebraic descriptions in Cartesian geometry.
But I found it surprisingly tricky to work out examples. Consider an ellipse with foci at (-1,0) and (+1,0), and define the set of points where the sum of its distances from these two is 4. Using Pythagoras’ theorem produces an equation with a bunch of square roots. Squaring both sides eventually turns this into
3x^2 + 4y^2 = 12
but this takes a bunch of steps of algebra, and mysterious cancellations of 4th-order terms.
Similarly, it takes a fair amount of work to get from the hyperbola with foci at (+2,+2) and (-2,-2), where the difference in distances is 4, to the simple equation
xy = 2.
Am I doing something wrong? Is there an easier way to get from foci and distances to these simple quadratic equations - without recourse to canonical forms, linear transformations, polar coordinates etc.?
Of course, I then want to talk about light waves bouncing from one focus to another… I’m not sure how to justify this without a little calculus.
- Cris _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun