Question

Graph the conic section.

Answer 1

\[25x^2-16y^2=400\]

Answer 2

First of all, we need to determine which of the conic sections this equation represents: ellipse, circle, parabola or hyperbola. I note that there's a negative sign (-) in that equation. Therefore, the conic section in question is a ... ??

Answer 3

hyperbola??

Answer 4

??

Answer 5

Sorry, Yana: I lost my connection.
Yes, you do have a hyperbola here.
Use the same general equation as before.
You must somehow transform the given equation into that standard form.

Answer 6

Standard form requires that you have a " 1 " on the right side of your equation. Therefore,
divide all terms by 100:\[25x^2-16y^2=400\]

Answer 7

I think everyone loses their connection...It needs and upgrade...ok...let me get this real quick:)

Answer 8

\[4x^2-6.25y^2=.25\]??

Answer 9

25/100 would come out to 1/4, wouldn't it?

Answer 10

yea.it would....my bad....

Answer 11

\[(25x^2-16y^2=400)~\div100\rightarrow x^2/4-y^2/(100/16)=400/400\]

Answer 12

wow...what???? that's alot...lol..anyways...

Answer 13

We end up with \[\frac{ x^2 }{4 }-\frac{ y^2 }{(100/16) }=1\]

Answer 14

If you think about this for a while, you'll probably see that a^2 = 4 and a =2; also (100/16)=(10/4)^2 or (5/2)^2, so that b = 5/2. Agreed or not?

Answer 15

Yes..I agree!:)(i really don't know what to say...so simple and easy!!:)) But you know I have to graph it right??

Answer 16

In summary:

a=2
b=5/2

Since that x^2 comes first and the y^2 second, we know that our hyperbola's graph is horizontal.

a=2 represents the distance from the center of the graph to the vertex.
b=5/2 is harder to explain; it has to do with the box we have to draw to draw the hyperbola.

There is an important relationship we need to know for the hyperbola" c^2=a^2+b^2.

c represents the distance from the center to one of the foci.

What is a^2? What is b^2? What is a^2 + b^2? Once you have a^2 + b^2, find the square root of the sum to obtain c, OK? BRB

Answer 17

Ok....cuz They gave me four graphs and they all look different...and dont know which ine to pick... I'll try to draw them..

Answer 18

Since I need to get off the 'Net pronto, I'll (this time) provide you with my proposed value of c: c=Sqrt(41)/2, or about 3.20.

Graph the following: (2,0), (-2,0), (3.20,0), (-3.20,0), and the lines y=5/2 and y=-5/2.

Answer 19

By doing this you will have shown the locations of the foci and vertices.