Question

Can somebody show me how to get these answers, I have the answers, but I don't know how they came to the answer.

Answer 1

center clear?

Answer 2

A.) (3, -5)
B.) 16
C.) (3,1) and (3,-11)
I don't know D.

Answer 3

i am only asking if it is clear how they got the center to be \((3,-5)\)

Answer 4

Yes.

Answer 5

oh so all is clear but the graph?

Answer 6

I don't understand how they worked out B and C and got the answer. And yes, D is confusing as well.

Answer 7

ok it is probably better to do D first

Answer 8

you have the center at \((3,-5)\) and the larger number is under the \(y\)
putting those facts together it has to look something like this

Answer 9

if the larger number was under the \(x\) then it would look like this

Answer 10

And, with the next steps, would yhu do something with the 16 and the other two coordinates? The major axis and the foci?

Answer 11

to get the foci, we need to find "\(c\)" using
\[c^2=a^2-b^2\]

Answer 12

general form of this is
\[\frac{(x-h)^2}{b^2}+\frac{(y-k)}{a^2}=1\] if \(a^2\) is the larger of the two numbers
in your case \(a^2=100,b^2=64\) so
\[a^2-b^2=100-64=36\]giving \(c=6\)

Answer 13

now it is important to know what it looks like
because we know the shape, it tell you that the foci are \(6\) units ABOVE AND BELOW the center, not to the left and right

Answer 14

... Wait, how did yhu get 6?

Answer 15

\[\frac{(x-h)^2}{b^2}+\frac{(y-k)}{a^2}=1\]
\[\frac{(x-3)^2}{64}+\frac{(y+5)}{100}=1\]\[a^2=100,b^2=64,c^2=a^2-b^2=100-64=36\]

Answer 16

since \(c^2=36\) you get \(c=6\)

Answer 17

Oh, Did yhu take the square root? . . .

Answer 18

kinda
i just know that \(6^2=36\) but if you want to say \(c=\sqrt{36}\) that is fine too

Answer 19

Oh, oh my goodness it's actually starting to make more sense. So then... 6 would be the radius... Right?

Answer 20

it doesn't really have a radius because it is not a circle
but 6 is the distance from the center to the foci
that is why it is important to know what it looks like, to know if you are supposed to go up and down from the center, or left and right

Answer 21

Oh. so... Would it be an oval like yhu drew, or would it be one of these... and the foci is where ever it's graphed?

Answer 22

no it is the oval i drew, not a parabola
like a flattened circle

Answer 23

now we know how to get the foci
6 units up form \((3,-5)\) is \((3,1)\) because \(-5+6=1\) and 6 units down is \((3,-11)\)

Answer 24

i hope those were the answers!

Answer 25

Oh!
So, is the circle that I draw, do the point's lie on the oval?

Answer 26

it is the oval that i drew above

Answer 27

the points do not lie on the oval, they are inside
the center is in the middle, and the two foci are also inside

Answer 28

Ok, so... Does it matter how big the oval is? Or no, as long as the points are inside?

Answer 29

no not really
unless you have graph paper

Answer 30

Oh my gosh. ^.^ Ok, do yhu mind if I draw out what I got and yhu can check to see if I got D right?

Answer 31

go ahead
it should look something like the one i drew above

Answer 32

plot the points \((3,-5)\) as the center and also plot the foci \((3,1)\) and \((3,-11)\) then draw the oval