- #1

chwala

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- 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...

##\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...