I would really appreciate it if someone would help me with the attached radical problem.
Thanks, you guys are always a great help.
I don't know how to show or paste here your attached computations, so let me just answer it.Originally Posted by coopsterdude
In the 1st line, y^2 is present in both sides of the equation, so you cancel it.
Then you either expand the (x+4)^2 and (x-4)^2, or use the a^2 -b^2 = (a+b)(a-b) method, to get 16x from those two perfect squares. Then the equation will become
16x = 144 -24sqrt[(x-4)^2 +y^2]
Isolate the radical,
16x -144 = -24sqrt[(x-4)^2 +y^2]
Divide both sides by 4, ------(why not by 8?),
4x -36 = -6sqrt[(x-4)^2 +y^2] ------------the 2nd line in your attachment.
The 3rd line, you can arrive at that by clearing the radical sign, square both sides of that 2nd line,
16x^2 -288x +1296 = 36[(x-4)^2 +y^2]
Then expand the righthand side, simplify, and you'd arrive at the last line in your attachment.
16x^2 -288x +1296 = 36[x^2 -8x +16 +y^2]
16x^2 -288x +1296 = 36x^2 -288x +576 +36y^2
You can cancel the two (-288x) since they are at both sides of the equation, or you can just transpose everything to the lefthand side. Or, since in the last line of your attachment, the variables are on the RHS while the constant is at the LHS, then isolate the constants in the LHS,
1296 -576 = 36x^2 -288x +36y^2 -16x^2 +288x
720 = 20x^2 +36y^2 ---------the last line in your attachment.
That is an ellipse.