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

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- Sep 13th 2010, 03:22 PMlvcneed help with hyperbola and ellipse
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 - Sep 13th 2010, 03:53 PMpickslides
$\displaystyle 5x^2 + 9y^2=45$

$\displaystyle \frac{5x^2}{5} + \frac{9y^2}{5}=\frac{45}{5}$

$\displaystyle x^2 + \frac{9y^2}{5}=9$

$\displaystyle \frac{x^2}{9} + \frac{9y^2}{9\times 5}=\frac{45}{9\times 5}$

$\displaystyle \frac{x^2}{9} + \frac{y^2}{ 5}=1$

$\displaystyle \frac{x^2}{\sqrt{9}^2} + \frac{y^2}{\sqrt{ 5}^2}=1$ - Sep 13th 2010, 06:58 PMSoroban
Hello, lvc!

Quote:

$\displaystyle 16x^2-4y^2+64\:=\:0$

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.)

- Sep 14th 2010, 03:00 AMHallsofIvy
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.