# need help with hyperbola and ellipse

• Sep 13th 2010, 04:22 PM
lvc
need 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, 04:53 PM
pickslides
$5x^2 + 9y^2=45$

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

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

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

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

$\frac{x^2}{\sqrt{9}^2} + \frac{y^2}{\sqrt{ 5}^2}=1$
• Sep 13th 2010, 07:58 PM
Soroban
Hello, lvc!

Quote:

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

We have: . $4y^2 - 16x^2 \:=\:64$

Divide by 64: . $\displaystyle \frac{4y^2}{64} - \frac{16x^2}{64} \;=\;\frac{64}{64} \quad\Rightarrow\quad \frac{y^2}{16} - \frac{x^2}{4} \;=\;1$

(In the first one, just divide by 45.)

• Sep 14th 2010, 04:00 AM
HallsofIvy
Quote:

Originally Posted by lvc
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

They are already in "hyerbola and ellipse form". What you mean is into "standard form".

Standard form for an ellipse is $\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$
and you have $5x^2+ 9x^2= 45$. Divide both sides by 45 to get "1" on the right.

Standard form for an ellipse is either $\frac{x^2}{a^2}- \frac{y^2}{b^2}= 1$ or $\frac{y^2}{b^2}- \frac{x^2}{a^2}= 1$. You have $16x^2- 4y^2+ 64= 0$ which is the same as $16x^2- 4y^2= -64$. (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.