How can I simplify my geometric calculation?

Question

Here is a math problem I am dealing with right now:

Given ellipse $C$: $x^2/a^2 + y^2/b^2=1$ ($a>b>0$). Ellipse $C$ passes through the point $P$: $(1,3/2)$, and has eccentricity $e=1/2$. Given line $\ell$: $x=4$. $\overline{AB}$ is the chord that intersects line $\ell$ on point $M$, and $F$ is the right focus of ellipse $C$. The slopes of $\overline{PA}$, $\overline{PB}$, and $\overline{PM}$ are respectively $k_1$, $k_2$, and $k_3$. Is there a constant $\lambda$ that satisfies $k_1 + k_2 =\lambda k_3$? If there is, please calculate $\lambda$.

How can use the last results into next input?

Solve[{x^2/4 + y^2/3 == 1, y == k (x - 1)}, {x, y}]

{x -> (2 (2 k^2 - 3 Sqrt[1 + k^2]))/(3 + 4 k^2), 
 y -> -k + (4 k^3)/(3 + 4 k^2) - (6 k Sqrt[1 + k^2])/(3 + 4 k^2)}, 
{x -> (2 (2 k^2 + 3 Sqrt[1 + k^2]))/(3 + 4 k^2), 
 y -> -k + (4 k^3)/(3 + 4 k^2) + (6 k Sqrt[1 + k^2])/(3 + 4 k^2)}}

A = 
  {(2 (2 k^2 - 3 Sqrt[1 + k^2]))/(3 + 4 k^2), 
   -k + (4 k^3)/(3 + 4 k^2) - (6 k Sqrt[1 + k^2])/(3 + 4 k^2)};
B = 
  {(2 (2 k^2 + 3 Sqrt[1 + k^2]))/(3 + 4 k^2), 
   -k + (4 k^3)/(3 + 4 k^2) + (6 k Sqrt[1 + k^2])/(3 + 4 k^2)};
M = {4, 3 k};
P = {1, 3/2};
Simplify[k1 = (A[[2]] - P[[2]])/(A[[1]] - P[[1]]), k ∈ Reals]
Simplify[k2 = (B[[2]] - P[[2]])/(B[[1]] - P[[1]])]
Simplify[k3 = (M[[2]] - P[[2]])/(M[[1]] - P[[1]])]
Simplify[k1 + k2]

(3 + 2 k + 4 k^2 + 4 k Sqrt[1 + k^2])/(2 + 4 Sqrt[1 + k^2])

(3 + 2 k + 4 k^2 - 4 k Sqrt[1 + k^2])/(2 - 4 Sqrt[1 + k^2])

-(1/2) + k

-1 + 2 k

How can I use the result of the last and put it in the next input?

Answer 1

M = {4, 3 k};
P = {1, 3/2};
{A, B} = Values@Solve[{x^2/4 + y^2/3 == 1, y == k (x - 1)}, {x, y}]
slope[A_, B_] := 1/Divide @@ (B - A);
{k1, k2, k3} = slope @@@ {{A, P}, {B, P}, {M, P}} // FullSimplify
FullSimplify[k1 + k2]