FORMULARIO DE GEOMETRIA ANALITICA
Enviado por James Rogers • 19 de Septiembre de 2022 • Apuntes • 1.466 Palabras (6 Páginas) • 108 Visitas
[pic 1] C(0, 0) x2 + y2 = r2 e = 0 | y2 = 4px e = 1 V(0, 0) F(p, 0) x = -p L.R. = |4p| | [pic 2] x2 = 4py e = 1 V(0, 0) F(0, p) y = - p L.R. = |4p| | C(0, 0) x 2 + y2 = [pic 3] [pic 4] 1 a2 b2 V’(-a, 0) V(a, 0) F’(-c, 0) F(c, 0) B’(0, -b) B(0, b) 2b2 LR = a a2 = b2 + c2 e = c < 1 a | [pic 5] C(0, 0) x2 y2 + = 1 b2 a2 V’(0, -a) V(0, a) F’(0, -c) F(0, c) B’(-b, 0) B(b, 0) 2b2 LR = a a2 = b2 + c2 e = c < 1 a |
[pic 6] C(h, k) e = 0 (x – h)2 + (y – k)2 = r2 x2 + y2 + Dx + Ey + F = 0 D E h = − k = − 2 2 r = 1 D2 + E2 − 4F 2 x2 ± bx = (x ± b )2 − (b )2 2 2 | [pic 7] v(h, k) (y – k)2 = 4p(x – h) F(h+p, k) x = h-p L.R. = |4p| e = 1 y2+Dx+Ey+F=0 ax ± b= a(𝒙 ± 𝒃) 𝒂 | v(h, k) (x – h)2 = 4p(y – k) F(h, k+p) y = k-p L.R. = |4p| e = 1 x2+Dx+Ey+F=0 | ||
[pic 8] C(h, k) (x − h)2 (y − k)2 + = 1 a2 b2 V’(h-a, k) V(h+a, k) F’(h-c, k) F(h+c, k) B’(h, k-b) B(h, k+b) 2b2 LR = a e = c < 1 a a2 = b2 + c2 Ax2+Cy2+Dx+Ey+F=0 | C(h, k) (x − h)2 + (y − k)2 = b2 a2 1 V’(h, k-a) V(h, k+a) F’(h, k-c) F(h, k+c) B’(h-b, k) B(h+b, k) 2b2 LR = a e = c < 1 a a2 = b2 + c2 Ax2+Cy2+Dx+Ey+F=0 | C(0, 0) x2 y2 − = 1 a2 b2 V’(-a, 0) V(a, 0) F’(-c, 0) F(c, 0) B’(0 –b) B(0, b) 2b2 LR = a c2 = a2 + b2 e = c > 1 a y = ± b x a Ax2+Cy2+F=0 | C(0, 0) y2 x2 − = 1 a2 b2 V’(0, -a) V(0, a) F’(0, -c) F(0, c) B’(-b, 0) B(b, 0) 2b2 LR = a c2 = a2 + b2 e = c > 1 a y = ± a x b Ax2+Cy2+F=0 | C(h, k) (x − h)2 − (y − k)2 = 1 a2 b2 V’(h-a, k) V(h+a, k) F’(h-c, k) F(h+c,k) B’(h, k-b) B(h, k+b) 2b2 LR = a c2 = a2 + b2 e = c > 1 a y = ± b (x-h)+k a Ax2+Cy2+Dx+Ey+F=0 |
d= P1P2= (x2 − x1)2 + (y2 − y1)2 x = x1 + rx2 y = y1 + ry2 r = x − x1 = y − y1 1 + r 1 + r x2 − x y2 − y | ||||
A = 1 [x1y2 + x2y3 + x3y1 − x1y3 − x3y2 − x2y1] 2 A = 1 [x1(y2 − yn) + x2(y3 − y1) + x3(y4 − y2). . .] 2 xcos(ω)+sen(ω)−p = 0 A x + B y + C = 0 r = A2 + B2 ± r ± r ± r cos(ω) = A sen(ω) = B − p = C ± r ± r ± r Ax + By + C = 0, P1(x1, y1) l1 || l2 d = Ax1 + By1 + C d = C1 − C2
± A2 + B2 A2 + B2 d = b2 − b1
A2 + B2 ± d1 = ±d2 ± A1x + B1y + C1 = ± A2 x + B2 y + C2 ± A2 + B2 ± A2 + B2 1 1 2 2 | C(h, k) (y − k)2 − (x − h)2 = a2 b2 1 V’(h, k-a) V(h, k+a) F’(h, k-c) F(h, k+c) B’(h-b, k) B(h+b, k) 2b2 LR = a c2 = a2 + b2 e = c > 1 a y = ± a (x-h)+k b Ax2+Cy2+Dx+Ey+F=0 | |||
x = x1 + x2 m 2 y + y ym = 1 2 2 x = x1 + r(x2 − x1) y = y1 + r(y2 − y1) r = x − x1 = y − y1 x2 − x1 y2 − y1 m = y2 − y1 x2 − x1 tanθ = m tan θ = m2 − m1 1 + m2m1 l1 || l2 ⇒ m1 = m2 l1 ⊥ l2 ⇒ m1m2 = −1 y − y1 = m(x − x1) y = mx + b y − y = y2 − y1 (x − x ) 1 x2 − x1 1 | x1 y1 x2 y2 A = 1 ⁝ ⁝ 2 x y n n x1 y1 Ax + By + C = 0 m = − A b = − C B B x + y = 1 a b a = − C b = − C A B l2 || l1 y − y1 = m1(x − x1) l2 ⊥ l1 y − y1 = − 1 (x − x1) m1 |
[pic 9][pic 10][pic 11][pic 12][pic 13][pic 14][pic 15][pic 16][pic 17][pic 18][pic 19][pic 20][pic 21][pic 22][pic 23][pic 24][pic 25][pic 26][pic 27][pic 28][pic 29][pic 30][pic 31][pic 32][pic 33][pic 34][pic 35][pic 36][pic 37][pic 38][pic 39][pic 40][pic 41][pic 42][pic 43][pic 44][pic 45][pic 46][pic 47][pic 48][pic 49][pic 50][pic 51][pic 52][pic 53][pic 54][pic 55][pic 56][pic 57][pic 58][pic 59][pic 60][pic 61]Academia de Matemáticas Turno Vespertino. Ing. Leobardo Sánchez Pimentel[pic 62][pic 63][pic 64][pic 65][pic 66][pic 67]
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