Question 1166142
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Since both equations have the same constant term, you can set the variable terms equal to each other thus and determine the ratio between *[tex \Large x] and *[tex \Large y], thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ +\ 9y^2\ =\ 9x^2\ +\ 4y^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -5x^2\ =\ -5y^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ y^2]


So the first thing we notice is that *[tex \Large x\ =\ \pm y] which means that the points of intersection are going to lie on the lines *[tex \Large y\ =\ x] and *[tex \Large y\ =\ -x], and that once we calculate one of the coordinates, we will have all of the others just by changing the signs.


Since *[tex \Large x^2\ =\ y^2], we can substitute into either of the equations


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ +\ 9x^2\ =\ 36]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \pm\frac{6}{\sqrt{13}}]


So the first point is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(\frac{6}{\sqrt{13}},\frac{6}{\sqrt{13}}\)]


And the other three are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(\frac{6}{\sqrt{13}},-\frac{6}{\sqrt{13}}\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(-\frac{6}{\sqrt{13}},-\frac{6}{\sqrt{13}}\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(-\frac{6}{\sqrt{13}},\frac{6}{\sqrt{13}}\)]


Your instructor may require you to rationalize your denominators, but I leave that as an exercise for you.

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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I > Ø
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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