# Math Help - More ellispe- need help

1. ## More ellispe- need help

Graph the ellipse.

((x-1)^2))/(9) + ((y-2)^2))/(4)=1

I got:

4(x+1)^2 + 9(y-2)^2=36
I divided all the terms by 36

then I got: ((x+1)^2))/(4) + ((y-2)^2)/(4)=1
the center is (-1,2)

This is where I am stuck at and dont know what to do next or how to graph it.

16(x-2)^2+4(y-1)^2=64
I divided all the terms by 64

then I got: ((x-2)^2)/(4) + ((y-1)^2)/(4)=1
the center is (2,1)

This is also where I get stuck.

Find the foci of the ellipse whose equation is given

((x+3)^2)/(36)+ ((y-3)^2)/(25)=1

I got: foci at (-3, +squareroot 11, 3) and (-3, -squareroot 11, 3)

2. Hello, soly_sol!

You're still making silly mistakes . . .

Graph the ellipse: . $\frac{(x-1)^2)}{9} + \frac{(y-2)^2}{4}\;=\;1$

I got:

. . . No!
The center is: . $(1,\,2)$

$4(x+1)^2 + 9(y-2)^2\:=\:36$

I divided all the terms by 36: . $\frac{(x+1)^2}{4} + \frac{(y-2)^2}{4}\;=\;1$ . . . . No!
$\frac{4(x+1)^2}{36} + \frac{9(y-2)^2}{36} \;=\;\frac{36}{36}\quad\Rightarrow\quad \frac{(x+1)^2}{9} + \frac{y-2)^2}{4} \;=\;1$
This is where I am stuck at and dont know what to do next or how to graph it.
If this is true, you need more help than we can give here.

$16(x-2)^2 + 4(y-1)^2\;=\;64$

I divided all the terms by 64: . $\frac{(x-2)^2}{4} + \frac{(y-1)^2}{4}\;=\;1$ . . . . Stop doing that!

Find the foci of the ellipse: . $\frac{(x+3)^2}{36} + \frac{(y-3)^2}{25} \;=\;1$

I got: foci at (-3, +squareroot 11, 3) and (-3, -squareroot 11, 3)
. . . . . This doesn't make sense
I assume you meant: . $\left(-3,\;3+\sqrt{11}\right),\;\;\left(-3,\;3-\sqrt{11}\right)$

3. 4(x+1)^2 + 9(y-2)^2=36
I divided all the terms by 36

I went back and corrected this one and this is the graph I got:

16(x-2)^2+4(y-1)^2=64
I divided all the terms by 64

I went back and corrected this one and this is the graph I got:

5. Originally Posted by soly_sol
(A) 4(x+1)^2 + 9(y-2)^2=36
I divided all the terms by 36

I went back and corrected this one and this is the graph I got:

(B) 16(x-2)^2+4(y-1)^2=64
I divided all the terms by 64

I went back and corrected this one and this is the graph I got:

Hello,

the lengthes(?) of the axes are OK but you forgot to calculate the coordinates of the centers of the ellipses:

(A) C(-1, 2)

(B) C(2, 1)

With your graphs the centers are allways at the origin.