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: . $9x^2\:=\:144-16y^2$

I got
. . . . no

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

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

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

The ellipse is 8 units wide and 6 units high.

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

I got
. . . . Good!

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

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

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

I got: . $\frac{x^2}{36} + \frac{y^2}{4} \;=\;1$ . . . . no

Foci at: . $(0,\:\pm4\sqrt{2})$
The ellipse is 4 units wide and 12 units high.
We have: . $a = 2,\;b = 6$

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

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 ?

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

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:
$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$
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