Elipsa x 2 a 2 + y 2 16 = 1 \frac{x^2}{a^2}+\frac{y^2}{16}=1 a 2 x 2 + 16 y 2 = 1 y 2 = 2 p x y^2=2px y 2 = 2 p x T ( 4 3 , 2 ) T(4\sqrt3,2) T ( 4 3 , 2 ) T F E F P TF_EF_P T F E F P F E F_E F E x x x F P F_P F P
Prikaži rješenje Rješenje Prvo odredimo nepoznate veličine za elipsu:
( 4 3 ) 2 a 2 + 4 16 = 1 , a 2 = 64 , e = 64 − 16 = 4 3 , F E ( 4 3 , 0 ) . \begin{gathered} \frac{(4\sqrt3)^2}{a^2}+\frac4{16}=1,\quad a^2=64,\\ e=\sqrt{64-16}=4\sqrt3,\quad F_E(4\sqrt3,0). \end{gathered} a 2 ( 4 3 ) 2 + 16 4 = 1 , a 2 = 64 , e = 64 − 16 = 4 3 , F E ( 4 3 , 0 ) . Zatim za parabolu:
4 = 2 p ⋅ 4 3 , p = 3 6 , F P ( 3 12 , 0 ) . 4=2p\cdot4\sqrt3,\quad p=\frac{\sqrt3}6,\quad F_P\left(\frac{\sqrt3}{12},0\right). 4 = 2 p ⋅ 4 3 , p = 6 3 , F P ( 12 3 , 0 ) . Površina trokuta je
P = 4 3 − 3 12 2 ⋅ 2 = 47 3 12 . P=\frac{4\sqrt3-\frac{\sqrt3}{12}}2\cdot2=\frac{47\sqrt3}{12}. P = 2 4 3 − 12 3 ⋅ 2 = 12 47 3 . 