已知m＞0 a＞0 直线l:x/a+y=m 与椭圆 x^2/a^2+y^2=1相切与P点

题目

已知m＞0 a＞0 直线l:x/a+y=m 与椭圆 x^2/a^2+y^2=1相切与P点
1、求实数m的值
2、设直线l‘:x/a+y=n 与椭圆I有两个不同的交点A,B若PA·PB的最小值为-1 求椭圆的长轴长.

答案

(1)
x/a + y = m,y = m - x/a
代入椭圆得2x² - 2max + a²(m² - 1) = 0 (i)
相切,判别式∆ = (-2ma)² - 4*2a²(m² - 1) = 4a²(2 - m²) = 0
a > 0,则m² = 2
m = √2 (舍去m = -√2 < 0)
(i)变为 2x² -2√2ax + a² = (√2x - a) = 0,x = a/√2 = √2a/2
P(√2a/2,√2/2)
(2)
x/a + y = n,y = n - x/a
代入椭圆得2x² - 2nax + a²(n² - 1) = 0
x₁ + x₂ = na
x₁x₂ = a²(n² - 1)/2
y₁ + y₂ = 2n - (x₁ + x₂)/a = 2n - n = n
y₁y₂ = (n - x₁/a)(n - x₂/a) = n² - n(x₁ + x₂)/a + x₁x₂/a² = n² - n² + (n² - 1)/2 = (n² - 1)/2
向量PA = (x₁ - √2a/2,y₁ - √2/2)
向量PB = (x₂ - √2a/2,y₂ - √2/2)
向量PA•向量PB = (x₁ - √2a/2)((x₂ - √2a/2) + (y₁ - √2/2)(y₂ - √2/2)
= x₁x₂ - (√2a/2)(x₁ + x₂) + a²/2 + y₁y₂ - (√2/2)(y₁ + y₂) + 1/2
= a²(n² - 1)/2 - (√2a/2)*na + a²/2 + (n² - 1)/2 - (√2/2)n + 1/2
= (1/2)(a² + 1)n² - (√2/2)(a² + 1)n
= [(a² + 1)/2](n² - √2n)
n = √2/2时,PA·PB取最小值-(a² + 1)/2 = -1
a² = 1,a = 1
长轴长2a = 2