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 $\endgroup$– csn899Jan 27 at 1:00
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1 Answer
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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 *)
