Tentukan koordinat titik – titik api ellips : 8x^2+16y^2-48x-32y-56=0

Jawab

8x² + 16y²- 48 x – 32 y – 56= 0
8x² – 48x  + 16y² – 32 y = 56
8(x²- 6x) + 16(y²-2y) = 56
8(x – 3)² + 16(y- 1)² = 56  + 8.3² + 16(-1)²
8(x – 3)²  + 16(y- 1)² = 56 + 72 + 16 
8(x – 3)² + 16(y-1)² = 144
(x – 3)²/16  + (y-1)²/8 = 1 → (x -p)²/a + (y-q)²/b² = 1

p= 3
q = 1
a²= 16 → a= 4
b² = 8 → b = 2√2
c²= a²+b²
c² =16+ 8 = 24
c = 2√6

titik ap  = titik fokus
F1 (p+c, q),
F2 p-c, q)

F1 =  {3+ 2√6) , 1)
F2 = (3 – 2√6, 1)