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1 of 5
𝑥 − 6 2
25
−
𝑦 + 2 2
4
= 1
La Hipérbola es un conjunto de puntos
en el plano (x, y) cuya diferencia a dos
puntos fijos llamados focos es constante.
Dadas las siguiente hipérbolas dar , las coordenadas del centro, de los
vértices, los focos
Aplicando la formula de focos
Focos: 𝑐2
= 𝑎2
+ 𝑏2
= 25+4 = 29
c= √29
f 1= ( 29 + 6, −2)
f 2= (− 29 − 6, −2)
Centro:
centro: (6,-2)
completar cuadrados para obtener la cónica en la forma canónica
𝑥2+𝑦2+6𝑦+2𝑦−15=0
Suma de términos semejantes
𝑥2+𝑦2+8𝑦−15=0
Se completa con el binomio cuadrado perfecto.
x + a 2
=𝑥2
+ 2𝑎𝑥 + 𝑎2
Se estructura según la ecuación de la hipérbola
𝑥2+𝑦2+8𝑦−15+31=31
𝑥2+𝑦2+8𝑦+16=31
𝑥2
31
+
(𝑦 + 4)2
31
=
31
31
𝑥2
31
+
(𝑦 + 4)2
31
= 1
Centro: (0, -4)
Focos: 𝑐2
= 𝑎2
+ 𝑏2
= 31+31=62
c= 62
f 1= ( 62, −4)
f 2= (− 62, −4)
Encontrar la ecuación canónica de una elipse cuyos vértices son respectivamente:
𝑣1=(−5;1),𝑣2=(5;1),𝑣3=(−1;1),𝑣4=(1;1)
𝑥 − 1 2
52
+
𝑥 − 1 2
12
Se aplica la ecuación de la elipse
𝑥−ℎ 2
a2 +
y−k 2
b2 = 1
Según los datos:
k: 1
a: 5
b: 1
h: 1
Remplazamos
Ecuación cónica:
𝑦 = 𝑚𝑥 + 𝑏
Pendiente: -5=m1
m2:
1
𝑚2
=
1
−5
Remplazamos en la ecuación
𝑦 = 𝑚
1
−5
+ 𝑏
5=
1
−5
(0)+b
5=b
Ecuación:
y=
1
−5
+5
Determine la ecuación de la recta perpendicular a y = -5x - 5 y que pasa por el punto (0, -5)

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Algebra 1.0

  • 1.
  • 2. 𝑥 − 6 2 25 − 𝑦 + 2 2 4 = 1 La Hipérbola es un conjunto de puntos en el plano (x, y) cuya diferencia a dos puntos fijos llamados focos es constante. Dadas las siguiente hipérbolas dar , las coordenadas del centro, de los vértices, los focos Aplicando la formula de focos Focos: 𝑐2 = 𝑎2 + 𝑏2 = 25+4 = 29 c= √29 f 1= ( 29 + 6, −2) f 2= (− 29 − 6, −2) Centro: centro: (6,-2)
  • 3. completar cuadrados para obtener la cónica en la forma canónica 𝑥2+𝑦2+6𝑦+2𝑦−15=0 Suma de términos semejantes 𝑥2+𝑦2+8𝑦−15=0 Se completa con el binomio cuadrado perfecto. x + a 2 =𝑥2 + 2𝑎𝑥 + 𝑎2 Se estructura según la ecuación de la hipérbola 𝑥2+𝑦2+8𝑦−15+31=31 𝑥2+𝑦2+8𝑦+16=31 𝑥2 31 + (𝑦 + 4)2 31 = 31 31 𝑥2 31 + (𝑦 + 4)2 31 = 1 Centro: (0, -4) Focos: 𝑐2 = 𝑎2 + 𝑏2 = 31+31=62 c= 62 f 1= ( 62, −4) f 2= (− 62, −4)
  • 4. Encontrar la ecuación canónica de una elipse cuyos vértices son respectivamente: 𝑣1=(−5;1),𝑣2=(5;1),𝑣3=(−1;1),𝑣4=(1;1) 𝑥 − 1 2 52 + 𝑥 − 1 2 12 Se aplica la ecuación de la elipse 𝑥−ℎ 2 a2 + y−k 2 b2 = 1 Según los datos: k: 1 a: 5 b: 1 h: 1 Remplazamos
  • 5. Ecuación cónica: 𝑦 = 𝑚𝑥 + 𝑏 Pendiente: -5=m1 m2: 1 𝑚2 = 1 −5 Remplazamos en la ecuación 𝑦 = 𝑚 1 −5 + 𝑏 5= 1 −5 (0)+b 5=b Ecuación: y= 1 −5 +5 Determine la ecuación de la recta perpendicular a y = -5x - 5 y que pasa por el punto (0, -5)