1. ## Ellipses

Graph the ellipse and locate the foci.

9x^2=144-16y^2

I got

x^2/5+y^2/9=1
I got

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

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

I got: x^2/36+y^2/4=1
foci at: (0,-4 squareroot 2) and (0, 4 squareroot 2)

2. Hello, soly_sol!

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

I got
. . . . 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.

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

I got
. . . . Good!

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

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

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

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

3. x^2/36+y^2/4=1

On this problem, I dont understand why it would be x^2/4 +y^2/36=1 ?

4. I dont understand how to do this problem. Can you please help me?

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

center at (-1, 3)

5. Originally Posted by soly_sol
I dont understand how to do this problem. Can you please help me?

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

center at (-1, 3)
Can you find the sizes of the semi-major and semi-minor axes from this diagram?

-Dan

6. Would the semi-major be (3,-3) and semi-minor be (-1,-6)?

7. 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

8. 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)

9. 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