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Solve[{x^2/a^2 - y^2/b^2 == 1, {x + c, y} . {x - c, y} == 0, a > 0, 
  b > 0, c > 0}, {x, y}]

The software keeps prompting for running, but cannot calculate the result. Why?

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    $\begingroup$ It gives a result almost instantly on Mathematica 13.2.1.0 but it's hugely complicated so I've put it on pastebin here $\endgroup$
    – flinty
    Commented May 13, 2023 at 12:27
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    $\begingroup$ sol = Solve[{x^2/a^2 - y^2/b^2 == 1, {x + c, y} . {x - c, y} == 0, c > a > 0, c > b > 0}, {x, y}, Reals] $\endgroup$
    – cvgmt
    Commented May 13, 2023 at 13:23
  • $\begingroup$ @csn899 This is an amplication and not a contradiction of cvgmt's comment's. I also am using Mathematica 13.2.1. Both Solve and Reduce complete quickly (1.4 seconds in Complex mode in Reduce on my ancient machine from 2011}. The ByteCount of the result is 60768 bytes. Timing[ByteCount[ sol = ToRadicals[ Reduce[{x^2/a^2 - y^2/b^2 == 1, {x + c, y} . {x - c, y} == 0, c > a > 0, c > b > 0}, {x, y}, Cubics -> True, Quartics -> True]]]] So what version of Mathematica are you using? $\endgroup$
    – anon
    Commented May 13, 2023 at 13:57

2 Answers 2

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$Version

(* "13.2.1 for Mac OS X ARM (64-bit) (January 27, 2023)"  *)

Clear["Global`*"]

Assuming[{a > 0, b > 0, c > 0}, 
 Solve[{x^2/a^2 - y^2/b^2 == 1, {x + c, y} . {x - c, y} == 0}, {x, y}] // 
  FullSimplify]

enter image description here

If you also restrict x and y to reals,

Assuming[{a > 0, b > 0, c > 0, Element[{x, y}, Reals]}, 
 Solve[{x^2/a^2 - y^2/b^2 == 1, {x + c, y} . {x - c, y} == 0}, {x, y}] // 
  FullSimplify]

enter image description here

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You can help Mathmatica solve the equations as follows:

ClearAll[a, b, c, x, y, X, Y, X2, Y2];
x = a*X; y = b*Y;
equs = Simplify[{x^2/a^2 - y^2/b^2 == 1, {x+c, y}.{x-c, y} == 0}];
Solve[ equs/. {X^2 -> X2, Y^2 -> Y2}, {X2, Y2}] //Simplify //InputForm
(* {{X2 -> (b^2 + c^2)/(a^2 + b^2), Y2 -> (-a^2 + c^2)/(a^2 + b^2)}} *)

where X2 == (x/a)^2, Y2 == (y/b)^2.

You asked

The software keeps prompting for running, but cannot calculate the result. Why?

I can not answer that question because it involves detailed knowledge of the actual software version of Mathematica and algorithms used by Mathematica. However, if you just want to get a useful solution to your equations, then I have given that to you using simple replacements before and after the Solve[].

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