Question

斜率为1的直线l与椭圆x^2/4+y^2=1相交于A,B两点，则AB最大值是多少

Answer 1

let A(x1,y1),B(x2,y2)
let equation of the line l be
y= x+c ( c is a constant) (1)
sub (1) into ellipse
x^2/4 + (x+c)^2= 1
5x^2+ 8cx + 4(c^2-1) =0
x1+x2 = -8c/5 (3)
x1x2= 4(c^2-1)/5 (4)
Similarly
(y-c)^2/4 + y^2 = 1
5y^2 - 2cy +(c^2-4) =0
y1+y2 = 2c/5 (5)
y1y2 = (c^2-4)/5 (6)

|AB|^2
=(x1-x2)^2+(y1-y2)^2
= (x1+x2)^2-4x1x2 +(y1+y2)^2 - 4y1y2
= (-8c/5)^2 - 16(c^2-1)/5 + (2c/5)^2 - 4(c^2-4)/5
= (1/5)(14c^2 - 16(c^2-1) - 4(c^2-4))
= (1/5)( 32-7c^2)
(|AB|^2)'
= (1/5)(-14c)
(|AB|^2)' =0
=> c=0
(|AB|^2)'' <0 ( max )
max |AB|^2 at c=0
=>max |AB| at c=0

ie y =x
x^2/4+x^2 =1
x^2 = 4/5
x = 2√5/5 or -2√5/5
when x = 2√5/5 , y = 2√5/5
when x = -2√5/5 , y = -2√5/5
max |AB| = √ (32/5) = 4√10/5