Tentukan persamaan jika di ketahui panjang suatu minor= 8 dan titik fokus(1,-6),dan(1,4)

Jawaban

• Jawaban diposting oleh: vara201

  Fokus1 (1,-6) ==> (p,q+c)
  fokus2 (1,4) ==> (p, q-c)
  
  q + c =-6
  q - c = 4
  -
  2c = -10
  c = -5
  
  q +c = -6
  q -5= - 6
  q = -1
  minor 8 :
  2b = 8
  b = 4
  a^2 = c^2-b^2
  = -5^2 - 4^2
  = 25 -16
  =9
  
  (y-q)^2 / a^2 - (x-p)^2 /b^2
  =(y-(-1)) ^2 / 9 - (x-1)^2 / 4^2
  =(y+1) ^2 /9 - (x-1)^2 / 16
  maap klo salah
• Jawaban diposting oleh: yasmin7718

  Fokus 1 (1,-6) ==> (p, q +c)
  fokus 2 (1,4) ==> (p, q-c)
  
  q +c = -6
  q-c = 4
  -
  2c = -10
  c = -5
  q+ c = -6
  q -5 = -6
  q = -1
  
  minor 8 :
  2b = 8
  b = 4
  a^2 = c^2 -b^2
  = -5^2 - 4^2
  = 25 -16
  = 9
  persamaan :
  (y-q) ^2 / a^2 - (x-p)^2 / b^2
  = (y-(-1)) ^2 / 9^2 - (x-1)^2 / 4^2
  = (y+1)^2 /81 - (x-1)^2 / 16
  maap klo salah