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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?

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    $\begingroup$ Can you perhaps explain what you're trying to do with a line and an ellipse? $\endgroup$ Nov 17, 2019 at 13:49
  • $\begingroup$ The line indicates that a focus is at (1,0). That does not seem correct. $\endgroup$ Nov 17, 2019 at 15:44
  • $\begingroup$ @J.M.willbebacksoon. It's a math problem I am dealing with right now. Ellipse C: x 2 /a 2 + y 2 /b 2 =1 (a>b>0). Ellipse C pass through point P (1,3/2), Eccentricity e=1/2. Line l: x=4. AB is the chord that intersect with line l on point M, And F is the right foci of ellipse C. The slopes of PA ,PB, PM are respectively k 1 , k 2 , k 3 . Is there a constant λ that meets k 1 + k 2 =λ k 3 ? If there is , please calculate λ $\endgroup$
    – kile
    Nov 18, 2019 at 2:56
  • $\begingroup$ Please include that description in your question. $\endgroup$ Nov 18, 2019 at 3:01
  • $\begingroup$ Alright. @J.M.willbebacksoon $\endgroup$
    – kile
    Nov 18, 2019 at 3:08

1 Answer 1

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enter image description here

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]

enter image description here

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