Question

Can someone help me with my conic sections assignment? I'm really confused.

Answer 1

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?

Answer 2

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

Answer 3

ellipse

Answer 4

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

Answer 5

\[\frac{ x^{2} }{ a ^{2}} + \frac{ y ^{2}}{ b ^{2} }\]

Answer 6

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 ?

Answer 7

(0,-2)

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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" ?

Answer 12

1?

Answer 13

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)

Answer 14

3?

Answer 15

yes

Answer 16

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

Answer 17

So is that the final answer?

Answer 18

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.