(1) Find the elliptic equation; (2) judging whether1| f1b |+1| f1d | is a fixed value, and if so, finding a fixed value;
(3) Make a straight line ⊥BD cross ellipse at A and C, and set the area of the quadrangle obtained by connecting four intersection points counterclockwise as S to find the value range of S. ..
For reference only, please see 1, 3 questions to ensure the correct thinking, because there are inevitably mistakes in the deduction process.
(1) Analysis: According to the definition of ellipse, 4a = 4 √ 3 and c= 1.
∴a=√3,b=√(3- 1)= √2
The elliptic equation is: x 2/3+y 2/2 =1.
(2) Analysis: Let B(x 1, y 1) and D(x2, y2).
When that slope of the straight line BD doe not exist
x 1 = x2 =- 1； y 1 =-2√3/3； y2=2√3/3
1/| f 1B |+ 1/| f 1D | = 2/(2√3/3)=√3
When that slope of the straight line BD exist
Let BD equation be y=k(x+ 1)
Substituting into the elliptic arrangement, we get (2+3k 2) x 2+6k 2x+3k 2-6 = 0.
Vieta theorem: x 1+x2 =-6k 2/(2+3k 2), x 1x2 = (3k 2-6)/(2+3k 2).
1/|f 1b|+ 1/|f 1d|= 1/√[(x 1+ 1)^2+y 1^2]+ 1/√[(x2+ 1)^2+y2^2]
= 1/√( 1+k^2)( 1/|x 1+ 1|+ 1/|x2+ 1|)
= 1/√( 1+k^2)(|x 1-x2|/|x 1x2+x 1+x2+ 1|
= 1/√( 1+k^2)*4√(3k^2+3)/4=√3
∴ 1/|f 1b|+ 1/|f 1d|=√3
③ Analysis: ∵AC⊥BD
|bd|=√[(x 1-x2)^2+(y 1-y2)^2]
When that slope of the straight line BD doe not exist
x 1 = x2 =- 1； y 1 =-2√3/3； y2=2√3/3
|BD|=y2-y 1=4√3/3
At this time |AC| is the long axis 2√3.
s = 1/2 | BD | * | AC | = 1/2 * 4√3/3 * 2√3 = 4
When that slope of the straight line BD exist
Let BD equation be y=k(x+ 1)
Substituting into the elliptic arrangement, we get (2+3k 2) x 2+6k 2x+3k 2-6 = 0.
Vieta theorem: x 1+x2 =-6k 2/(2+3k 2), x 1x2 = (3k 2-6)/(2+3k 2).
(x 1-x2)^2=( x 1+x2)^2-4x 1x2=48(k^2+ 1)/(2+3k^2)^2
(y 1-y2)^2=k^2(x 1-x2)^2
|bd|^2=( 1+k^2)(x 1-x2)^2
= = & gt|bd|=√( 1+k^2)* 4√[3(k^2+ 1)]/(2+3k^2)=4√3(k^2+ 1)/(2+3k^2)
The AC equation is y=- 1/k(x- 1).
Substituting into the ellipse gives (2+3/k 2) x 2-6/k 2x+3/k 2-6 = 0.
= = & gt(2k^2+3)x^2-6x+3-6k^2=0
Vieta theorem: x3+x4 = 6/(2k 2+3), x3x4 = (3-6k 2)/(2k 2+3).
(x3-x4)^2=( x3+x4)^2-4x3x4=48k^2(k^2+ 1)/(2k^2+3)^2
(y3-y4)^2=(x3-x4)^2/k^2
|ac|^2=( 1+ 1/k^2)(x3-x4)^2
= = & gt|ac|=√( 1+ 1/k^2)* 4k√[3(k^2+ 1)]/(2k^2+3)=4√3(k^2+ 1)/(2k^2+3)
s= 1/2|bd|*|ac|= 1/2*4√3(k^2+ 1)/(2+3k^2)* 4√3(k^2+ 1)/(2k^2+3)
=24(k^2+ 1)^2/(6k^4+ 13k^2+6)
=24/(6+k^2/( 1+k^2)^2)
When k = 1, the minimum value of s is 96/25.
When k=0 or does not exist, s takes the maximum value of 4.