i have the eqautions
5x^2 + 9y^2=45
16x^2-4y^2+64=0
how do i make these into the hyperbola and ellipse form
$\displaystyle 5x^2 + 9y^2=45$
$\displaystyle \frac{5x^2}{5} + \frac{9y^2}{5}=\frac{45}{5}$
$\displaystyle x^2 + \frac{9y^2}{5}=9$
$\displaystyle \frac{x^2}{9} + \frac{9y^2}{9\times 5}=\frac{45}{9\times 5}$
$\displaystyle \frac{x^2}{9} + \frac{y^2}{ 5}=1$
$\displaystyle \frac{x^2}{\sqrt{9}^2} + \frac{y^2}{\sqrt{ 5}^2}=1$
Hello, lvc!
$\displaystyle 16x^2-4y^2+64\:=\:0$
We have: .$\displaystyle 4y^2 - 16x^2 \:=\:64$
Divide by 64: .$\displaystyle \displaystyle \frac{4y^2}{64} - \frac{16x^2}{64} \;=\;\frac{64}{64} \quad\Rightarrow\quad \frac{y^2}{16} - \frac{x^2}{4} \;=\;1$
(In the first one, just divide by 45.)
They are already in "hyerbola and ellipse form". What you mean is into "standard form".
Standard form for an ellipse is $\displaystyle \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$
and you have $\displaystyle 5x^2+ 9x^2= 45$. Divide both sides by 45 to get "1" on the right.
Standard form for an ellipse is either $\displaystyle \frac{x^2}{a^2}- \frac{y^2}{b^2}= 1$ or $\displaystyle \frac{y^2}{b^2}- \frac{x^2}{a^2}= 1$. You have $\displaystyle 16x^2- 4y^2+ 64= 0$ which is the same as $\displaystyle 16x^2- 4y^2= -64$. (We cannot divide "0" by anything to get "1" so I subtracted 64 from both sides.) Divide both sides by -64 to get "1" on the right.