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## anonymous one year ago Can someone help me with my conic sections assignment? I'm really confused.

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1. anonymous

I have a few that I really don't know how to do, but here's one: What is the equation of the following graph?

2. phi

First, what shape is that? is it a circle, ellipse, parabola or hyperbola?

3. anonymous

ellipse

4. phi

next, look in your notes for the equation of an ellipse. can you write down the "general form" ?

5. anonymous

$\frac{ x^{2} }{ a ^{2}} + \frac{ y ^{2}}{ b ^{2} }$

6. phi

an equation has an = sign in it. it should be $\frac{ x^{2} }{ a ^{2}} + \frac{ y ^{2}}{ b ^{2} }=1$ but that is for an ellipse whose center is at (0,0) your ellipse is not centered at (0,0). The better equation is $\frac{ (x-h)^{2} }{ a ^{2}} + \frac{ (y-k) ^{2}}{ b ^{2} } =1$ where (h,k) is the center of the ellipse. Can you figure out what point is in the exact center of your ellipse ?

7. anonymous

(0,-2)

8. phi

that means you know (h,k) h=0 and k= -2 put those into the equation

9. anonymous

$\frac{(x-0)^{2} }{ a ^{2} }+ \frac{ (y+2)^{2} }{ b ^{2}}$

10. phi

= 1 don't leave that out also, x-0 simplifies to x so $\frac{x^{2} }{ a ^{2} }+ \frac{ (y+2)^{2} }{ b ^{2}} =1$

11. phi

now for the "a" value: that is the distance in the x-direction (sideways) from the center to the side. what is the value of "a" ?

12. anonymous

1?

13. phi

yes, a is 1 and a^2 which means a*a is 1*1 = 1 so $\frac{x^{2} }{ a ^{2} }+ \frac{ (y+2)^{2} }{ b ^{2}} =1 \\ \frac{x^{2} }{ 1 ^{2} }+ \frac{ (y+2)^{2} }{ b ^{2}} =1$ now find b, which the distance from the center to the top (i.e. in the up/down direction , or y-axis direction)

14. anonymous

3?

15. phi

yes

16. phi

$\frac{x^{2} }{ 1 ^{2} }+ \frac{ (y+2)^{2} }{ 3^{2}} =1$ we could simplify that a little, and write $x^2 + \frac{ (y+2)^{2} }{ 9} =1$ but either way is ok

17. anonymous

So is that the final answer?

18. phi

yes. You can check some points to make sure it's correct. for example , when x is 0 you get $0^2 + \frac{ (y+2)^{2} }{ 9} =1 \\ \frac{ (y+2)^{2} }{ 9} =1 \\ (y+2)^2= 9 \\ y+2= \pm 3$ and y= +3-2= 1 or y= -3-2= -5 so points (0,1) and (0, -5) should be on the ellipse.

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