• October 25th 2006, 08:36 PM
jhonwashington
Hello everybody, how are you doing

I have a homework, I already did it almost all, but there are a few exercises which are so hard to me, could you please give me a hand

one exercise says

Find the vertices and foci of this ellipse

x^2/2+y^2/4=1

another exercise says

find the center, foci and vertices of the given ellipse

2x^2+3y^2-8x+6y+5=0

And the last one, sorry!

find the equation for the hyperbola described

Center (0,0) focus at (3,0) vertex at (1,0)

Thanks again.
• October 25th 2006, 10:42 PM
earboth
Quote:

Originally Posted by jhonwashington
...
Find the vertices and foci of this ellipse
x^2/2+y^2/4=1
...

Hi,

the axis of the ellipse lie on the coordinate-axis. The center of the ellipse is the origin of the coordinate system. The major axis a is situated on the y-axis, the minor axis b lies on the x-axis.
the length of the major axis is: $l_{a}=\sqrt{4}=2$
the length of the minor axis is: $l_{b}=\sqrt{2}$

The foci lie on the y-axis. Let e be the distance of a focus from the center, then you calculate e by:
$e^2=a^2-b^2$

Plug in the values you know. You get $F_1(0, \sqrt{2}) \text{ and } F_2(0, -\sqrt{2})$

Oops, sorry, I forgot to calculate the vertices:
You already know the length of axis. So I think it is no problem for you to get the coordinates of the vertices. For confirmation only: $[(0,2),\ (0, -2),\ (\sqrt{2} ,0),\ (-\sqrt{2}, 0)]$

EB
• October 25th 2006, 10:59 PM
earboth
Quote:

Originally Posted by jhonwashington
...
find the center, foci and vertices of the given ellipse
2x^2+3y^2-8x+6y+5=0
...

Hi,

the general equation of an ellipse is:
$\frac{(x-c)^2}{a^2} +\frac{(y-d)^2}{b^2}=1$
where M(c, d) is the center of the ellipse and a and b are the axis of the ellipse.

You only have to transform the given equation into the general form of the ellipse:

$2x^2+3y^2-8x+6y+5=0$
$2(x^2-4x)+3(y^2+2y)=-5$. Complete the squares:
$2(x^2-4x+4)+3(y^2+2y+1)=-5+8+3$. I've added 4 in the first bracket to get a complete square, so I added 8 at the LHS of the equation. To keep the balance you have to add 8 on the RHS too!
$2(x-2)^2+3(y+1)^2=6$. Divide by 6 (Remember the RHS of the general equation is 1)
$\frac{(x-2)^2}{3}+\frac{y+1)^2}{2}=1$

Now you know the length of the axis, the coordinates of the center (M(2,-1)). All further calculations are similar to the first problem. So I leave them to you.

EB
• October 25th 2006, 11:13 PM
earboth
Quote:

Originally Posted by jhonwashington
...
find the equation for the hyperbola described
Center (0,0) focus at (3,0) vertex at (1,0)
...

Hi,

the general equation of the described hyperbola is:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

You know by the vertex the length of axis a = 1.

You know that the coordinate of a focus F(e,0) is calculated by: $e^2=a^2+b^2$

Plug in the values you know and you'll get $b^2 = 8$

Now screw all parts together and you'll come up with:
$\frac{x^2}{1}-\frac{y^2}{8}=1$

EB
• October 26th 2006, 09:40 AM
jhonwashington
THANKs
thank you so much for your help earboth
• December 7th 2006, 02:39 PM
jhonwashington
how are you earboth

I just have a doubt about the exercise 2x^2+3y^2-8x+6y+5=0

you put as answer to this exercise
(x-2)^2 + y+1^2 = 1
_______ _____________
3 2

I just don't get how you came up with those denominators 3 and 2? could you please explain me why

thank you.
• December 7th 2006, 02:43 PM
jhonwashington
Ok I couldn't posted it correctly, I hope you understand what I'm saying

I tried to copy the answer just as you did, but I couldn't, that line which doesn't have denominator is supposed to have a 2. and the left line just a 3 as denominator.
• December 7th 2006, 06:18 PM
topsquark
Quote:

Originally Posted by jhonwashington
Ok I couldn't posted it correctly, I hope you understand what I'm saying

I tried to copy the answer just as you did, but I couldn't, that line which doesn't have denominator is supposed to have a 2. and the left line just a 3 as denominator.

Quote:

Originally Posted by earboth
$2(x-2)^2+3(y+1)^2=6$

$\frac{2(x-2)^2}{6} + \frac{3(y+1)^2}{6}=1$

$\frac{(x-2)^2}{3} + \frac{(y+1)^2}{2}=1$

-Dan