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Assume that the equation is:

$a^2 y_0 -a^2 y_0 (y y_0)/b^2 =x y_0 x_0$

The form of the output equation is as follows: the right side of the equation equal sign is 1.

The desired result of the integral equation is:

$x x_0/a^2+y y_0/b^2 = 1$

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  • $\begingroup$ sorry! I entered an extra plus sign $\endgroup$
    – csn899
    Commented Jan 27, 2023 at 1:00

1 Answer 1

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$Version

(* "13.2.0 for Mac OS X x86 (64-bit) (November 18, 2022)" *)

Clear["Global`*"]

eqn1 = a^2 y0 - a^2 y0 (y y0)/b^2 == x y0 x0;

eqn2 = SubtractSides[eqn1, eqn1[[1, 2]]]

(* a^2 y0 == x x0 y0 + (a^2 y y0^2)/b^2 *)

eqn3 = Assuming[eqn2[[1]] != 0,
  Simplify /@ DivideSides[eqn2, eqn2[[1]]]]

(* 1 == (x x0)/a^2 + (y y0)/b^2 *)

eqn4 = Reverse@eqn3

(* (x x0)/a^2 + (y y0)/b^2 == 1 *)
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