1. ## graphing problems

Need Help With This Math College Algebra

Attachment 1858

if you can download and view PDF files let me know and i will write out the questions here, the only reason i have it on pdf is because there is a graph along with it. ANY HELP IS GREATLY APRECIATED!

2. Originally Posted by JJbong
Need Help With This Math College Algebra

Attachment 1858

if you can download and view PDF files let me know and i will write out the questions here, the only reason i have it on pdf is because there is a graph along with it. ANY HELP IS GREATLY APRECIATED!

it says the attachment is invalid, i cant open it

3. ## i will type the question

hey i will type the question and it will be up in 5 minutes

4. ## Question Typed Out

1. Write the equation of the parabola given the focus at (-3,2) and the directrix is the line y= -6. It is helpful to draw a sketch of what is given to help in choosing the correct formula. Be sure you do have the correct formula to fill in. Leave the equation in standard form.

2. Given the ellipse equation 9x^2 + 4y^2 - 18x + 16y - 11=0, find the coordinates of the center, vertices, and the foci. Sketch the ellipse on the grid showing each of these points as dots. Shoe steps for converting to standard form.

I guess for the part whre you are supposed to plot the points......just give me the coordinates and any other necessary info on how to sketch myself.

THANK YOU

5. Originally Posted by JJbong
2. Given the ellipse equation 9x^2 + 4y^2 - 18x + 16y - 11=0, find the coordinates of the center, vertices, and the foci. Sketch the ellipse on the grid showing each of these points as dots. Shoe steps for converting to standard form.
9x^2+4y^2-18x+16y-11=0

9(x^2-2x)+4(y^2+4)=11

9(x^2-2x+1)+4(y^2+4+4) = 11+9+16=36

9(x-1)^2+4(y+2)^2=6^2

This is an ellipse with center at (1,-2).

6. Originally Posted by JJbong
1. Write the equation of the parabola given the focus at (-3,2) and the directrix is the line y= -6. It is helpful to draw a sketch of what is given to help in choosing the correct formula. Be sure you do have the correct formula to fill in. Leave the equation in standard form....

Hello,

maybe this is too late but here comes #1:

The focus is above the directrix thus it is a parabola which opens upward.
The vertex of the parabola is the midpoint between directrix and focus on a perpendicular line to the directrix. Thus V(-3, -2)

p is the distance beween directrix and focus; here p = 8
V(a, b)

then the general equation of the paraboal is:

2p(y - b) = (x - a)²

Plug in the values you know:

2*8*(y + 2) = (x + 3)²
Expand the brackets and collect all terms but the y on the RHS:

y = 1/16*x² + 3/8*x - 23/16

I've attached a diagram of the parabola.

EB