# Ellipses

• Nov 22nd 2007, 12:34 PM
soly_sol
Ellipses

Graph the ellipse and locate the foci.

9x^2=144-16y^2

I got
http://i48.photobucket.com/albums/f2...sol/f1q2g8.jpg

x^2/5+y^2/9=1
I got
http://i48.photobucket.com/albums/f2...l/f1q3g5-1.jpg

Find the standard form of the equation of the ellipse and give the location of its foci.

http://i48.photobucket.com/albums/f2...sol/f1q4g1.jpg

I got: x^2/9 +y^2/4=1
foci at: (- squareroot of 5,0) and (squareroot 5,0)

http://i48.photobucket.com/albums/f2...sol/f1q5g1.jpg
I got: x^2/36+y^2/4=1
foci at: (0,-4 squareroot 2) and (0, 4 squareroot 2)
• Nov 22nd 2007, 01:13 PM
Soroban
Hello, soly_sol!

Quote:

Graph the ellipse and locate the foci: .$\displaystyle 9x^2\:=\:144-16y^2$

I got
http://i48.photobucket.com/albums/f2...sol/f1q2g8.jpg . . . . no

We have: .$\displaystyle 9x^2 + 16y^2 \:=\:144$

Divide by 144: .$\displaystyle \frac{x^2}{16} + \frac{y^2}{9} \;=\;1$

Hence: .$\displaystyle a = 4,\;b = 3$

The ellipse is 8 units wide and 6 units high.

Quote:

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

I got
http://i48.photobucket.com/albums/f2...l/f1q3g5-1.jpg . . . . Good!

Quote:

Find the standard form of the equation of the ellipse and give the location of its foci.

http://i48.photobucket.com/albums/f2...sol/f1q4g1.jpg

I got: .$\displaystyle \frac{x^2}{9} + \frac{y^2}{4} \;=\;1$ . . . . Yes!

Foci at: .$\displaystyle (\pm\sqrt{5},\:0)$ . . . . Right!

Quote:

http://i48.photobucket.com/albums/f2...sol/f1q5g1.jpg
I got: .$\displaystyle \frac{x^2}{36} + \frac{y^2}{4} \;=\;1$ . . . . no

Foci at: .$\displaystyle (0,\:\pm4\sqrt{2})$

The ellipse is 4 units wide and 12 units high.
We have: .$\displaystyle a = 2,\;b = 6$

The equation is: .$\displaystyle \frac{x^2}{4} + \frac{y^2}{36} \;=\;1$

• Nov 22nd 2007, 03:56 PM
soly_sol
x^2/36+y^2/4=1

On this problem, I dont understand why it would be x^2/4 +y^2/36=1 ?
• Nov 22nd 2007, 04:19 PM
soly_sol

Find the standard form of the equation of the ellipse and give the location of its foci.

http://i48.photobucket.com/albums/f2...sol/f1q6g1.jpg
center at (-1, 3)
• Nov 23rd 2007, 05:20 AM
topsquark
Quote:

Originally Posted by soly_sol

Find the standard form of the equation of the ellipse and give the location of its foci.

http://i48.photobucket.com/albums/f2...sol/f1q6g1.jpg
center at (-1, 3)

Can you find the sizes of the semi-major and semi-minor axes from this diagram?

-Dan
• Nov 23rd 2007, 11:23 AM
soly_sol
Would the semi-major be (3,-3) and semi-minor be (-1,-6)?
• Nov 24th 2007, 06:37 AM
topsquark
Quote:

Originally Posted by soly_sol
Would the semi-major be (3,-3) and semi-minor be (-1,-6)?

Huh? The semi-major and minor axes are distances. Just like the radius of a circle. The semi-major axis (a) is half of the "long axis" of the ellipse and the semi-minor axis (b) is half of the "short axis." So I'm getting a = 4 and along the x direction and b = 3 along the y direction. So the form for your ellipse will be:
$\displaystyle \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$
for this form of ellipse, where (h, k) is the center point of the ellipse.

-Dan
• Nov 24th 2007, 11:22 AM
soly_sol
Can you please explain to me how to get the semi-major and semi-minor?

Also, this is the answer I got:

((x+1)^2)/(16) + ((y+3)^2)/(9)
foci: (-1 + squareroot 7, -3) and (-1 negative squareroot 7,-3)
• Nov 24th 2007, 02:40 PM
topsquark
Quote:

Originally Posted by soly_sol
Can you please explain to me how to get the semi-major and semi-minor?

If you are having this many problems about how to do this you really need to speak with your teacher for extra help.

I simply looked at the ellipse, and noted it was wider than it was tall. So the horizontal axis is the major axis. Then I noted the ellipse is 8 units across, so the semi-major axis has a length of 4. Similarly for the vertical axis.

-Dan