Solve the given simultaneous equations that involves hyperbolas

[chwala]

Homework Statement

This is my own question- i am currently reading on ellipses and hyperbolas..

$\dfrac{x^2}{4}-\dfrac{y^2}{9}=1$

$\dfrac{y^2}{4}-\dfrac{x^2}{9}=1$

Relevant Equations

Simultaneous equations

My approach on this;

$\dfrac{x^2}{4}-\dfrac{y^2}{9}=\dfrac{y^2}{4}-\dfrac{x^2}{9}$

$9x^2-4y^2=9y^2-4x^2$

$13x^2-13y^2=0$

$x^2=y^2$

Therefore, on substituting back into equation we shall have;

$\dfrac{x^2}{4}-\dfrac{x^2}{9}=1$

$9x^2-4x^2=36$

$5x^2=36$

$x^2=7.2$

$x=\sqrt{7.2}=±2.68$ to 3 significant figures.

therefore our solutions are; $(x,y)= (2.68, 2.68), (2.68,-2.68), (-2.68, -2.68), (-2.68,2.68).$

I would appreciate any other approach other than this...

 

[fresh_42]

Homework Statement:: This is my own question- i am currently reading on ellipses and hyperbolas..

$\dfrac{x^2}{4}-\dfrac{y^2}{9}=1$

$\dfrac{y^2}{4}-\dfrac{x^2}{9}=1$
Relevant Equations:: Simultaneous equations

I would appreciate any other approach other than this...

Why? You have a nice solution.

 

[chwala]

Why? You have a nice solution.

Cheers @fresh_42 ...different ways in solving math problems has always fascinated me...other than finding the solution...I know that this forum has the smartest brains on the planet... another approach may be forthcoming ... ...cheers mate!

I guess since $x^2 = y^2$, not much can be done here...

 

[fresh_42]

However you look at it, you have to calculate the intersection points of two conic sections.

You can only get $x^2=y^2$ directly by the observation that $(x^2,y^2)=(y^2,x^2)$ for symmetry reasons, and if $x\neq \pm y$ then we would get more than four intersection points which is impossible. So $9x^2-4x^2=5x^2=36$ and thus $x=\pm \sqrt{7.2}$ follows without many calculations.

 

[Mark44]

Nit in the second line below:

$x^2=7.2$
$x=\sqrt{7.2}=±2.68$ to 3 significant figures.

That line should start with $x = \pm \sqrt{7.2} \dots$

If you are thinking that $\sqrt{7.2} =± 2.68...$, that is erroneous thinking.