According to a focus f (2,0), we can get: a-b = 2 = 4 ①.

Then the two directrix of the ellipse are: x = a/2.

∴ The distance between two directrix is 2 * (a/2) = λ.

& lt= & gta^=λ

& lt= & gtb^=a^-4=λ-4

The elliptic equation is: x/λ+y/(λ-4) = 1.

2. Let the symmetry point of F about L be B(x 1, y 1).

According to the meaning of symmetry, the line segment FB is vertically bisected by the straight line L.

Let FB and l intersect at p, then p must be the midpoint of FB, and l⊥FB.

Let the slope of the straight line L be k, then: kfb =-1/k1=-1/k2.

And FB must pass f (2,0).

According to the point tilt formula, kFB=- 1/k, f (2,0), the equation of FB can be obtained as follows:

FB:y=(- 1/k)*(x-2)

When the straight line L passes through A( 1, 0), the following equation can be obtained according to the point inclination formula:

l:y=k(x- 1)

Combining the equations of FB and L, we can get the coordinate p of their intersection point as follows:

p((k^+2)/(k^+ 1),k/(k^+ 1))

It has been proved that P is the midpoint of FB, then B(x 1, y 1) can be obtained according to the midpoint coordinate formula:

x 1=2*xP-xF

y 1=2*yP-yF

Substituting the coordinates of points p and f, we can get:

x 1=2/(k^+ 1)

y 1=2k/(k^+ 1)

Namely b (2/(k+ 1), 2k/(k+ 1))

And point b is on the ellipse according to the meaning of the question, and it is brought into the elliptic equation solved in the first question, and sorted out, so as to get a quadratic equation with one variable (including λ) about k:

(λ^-4λ)*(k^)^+(2λ^- 12λ)*k^+(λ-4)^=0

The equation must have real roots, so there are:

△=(2λ^- 12λ)^-4*(λ^-4λ)*(λ-4)^≥0

& lt= & gtλ≤ 16/3

And the equation is about k, k ≥ 0, and the two real roots of the equation must be non-negative, then there are:

Two sums: -(2λ- 12λ)/(λ-4λ) ≥ 0.

Two products: (λ-4)/(λ-4 λ) ≥ 0

Binding condition λ > 4. Available quantity: 4

Combining with ③ formula, we can get the range of λ:

λ∈(4, 16/3]