Question

解析几何 椭圆的几道题

解析几何 椭圆的几道题1.已知过点P(1,1)引一弦与椭圆x平方＋2y平方＝4交于AB两点，使丨PA丨=丨PB丨 求AB所在的直线方程及AB的长。
2.已知椭圆的标准方程x平方/16+y平方/12=1 设O是原点，直线y=x+m与椭圆交于M N两点 当三角形OMN的面积最大时，求m的值

Answer 1

1.
令AB的斜率为k， 方程为y - 1 = k(x - 1), y = kx + 1 - k
带入椭圆: x² + 2(kx + 1 - k)² - 4 = 0, (2k² + 1)x² +4k(1-k)x + 2(1 - k)² - 4 = 0 (i)
令A(p, p'), B(q, q'), 则p + q = 4k(k - 1)/(2k² + 1)
AB的中点P的横坐标为1 = (p + q)/2 = 2k(k - 1)/(2k² + 1)
k = -1/2
直线为y = -x/2 + 3/2
(i)变为 3x² - 6x + 1 = 0
p+q = 2, pq = 1/3
|AB| = √[(p - q)² + (p' - q')²] = √[(p - q)² + (-p/2 + 3/2 + q/2 - 3/2)²] = √5/2[(p + q)² - 4pq] = √5/2(4 - 4/3) = 4√5/3

2.
令M(p, p + m), N(q, q + m)
将直线带入椭圆x²/16 + y²/12 = 1, x²/12 + 4(x + m)²/12 - 1 = 0
7x² + 8mx + 4(m² - 12) = 0
p+q = -8m/7, pq = 4(m² - 12)/7
|MN|² = (p - q)² + (p +m - q - m)² = 2[(p + q)² - 4pq] = 2[(-8m/7)² - 16(m² - 12)/7] = (-96/49)(m² - 28)
原点与MN (x - y + m = 0)的距离为d = |m|/√2
S² = (1/4)|MN|²*d² = (12/49)m²(28 - m²)
m² = 28 - m², m = ±√14时, S²最大, S最大