证明：
|PF1|²
=(x - c)² + y²
=[a²(x - c)² + a²y²]/a²
=[a²x² - 2a²cx + a²c² + a²y²]/a² /***--根据b²x² + a²y² = a²b² ***/
=[a²x² - 2a²cx + a²c² + a²b² - b²x²]/a²
=[(a²-b²)x² = 2a²cx + a²(b² + c²)]/a²
=[c²x² -2a²cx + a^4]/a²
=(a² - cx)²/a²
∴PF1 = (a² - cx)/a = a - (c/a)x = a - ex
同理可证：PF2 = a + ex