Question

Can somebody explain to me how to graph Conic Sections and how to find the center, foci , and all of that from an equation and how to take for example "16x^2 -81y^2 + 324 y – 1620 = 0" into an equation we can graph... just everything.. I need help <3 My teacher isnt helping at all. ;-;

Answer 1

So are you familiar with the equations of each type of conic section?

Answer 2

... no....

Answer 3

hey

Answer 4

hiya, can you help me.....? ;-;

Answer 5

sure

Answer 6

You're the best. omg

Answer 7

just let me know

Answer 8

Have you looked at this? http://en.wikipedia.org/wiki/Conic_section#Features

Answer 9

I havent..

Answer 10

Look especially at this section - http://en.wikipedia.org/wiki/Conic_section#Cartesian_coordinates

Answer 11

one sec

Answer 12

So the equation up top is a circle? where does the y come from?

Answer 13

324y*

Answer 14

or is it a parabola ?

Answer 15

So we have
$$
16x^2 -81y^2 + 324 y – 1620 = 0
$$

$$

Ax^2 + Bxy + Cy^2 +Dx + Ey + F = 0

$$

What is A-F?

Then use
$$
B^2-4AC
$$

To determine which type of conic section using the last link I sent.

In other words, is the last equation, <,> or equal to zero?

Answer 16

Do you see where the "y" comes from now?
For example, A=16. See that?

Answer 17

BRB, figure out the rest -- I'll check your work.

Answer 18

Yea, But what about the parts that aren't there like Bzy and Dz?

Answer 19

Those would be zero then. BRB

Answer 20

Okay... so B=0? A=16 and C=-81?
so 0^2-4(16*-81)
=5184
5184>0
so it'd be a hyperbola?

Answer 21

In my assignment i need this for i need to graph the points and the equation.
for example: Find the graphing form, coordinates of the center & foci, the length of the transverse and conjugate axis, then graph the points and equation.
16x2 -81y2 + 324 y – 1620 = 0

Answer 22

yes, hyperbola

Answer 23

So we now think its a hyperbola, what is the equation of a hyperbola? According to the 1st link I sent what is the equation - http://en.wikipedia.org/wiki/Conic_section#Features

Answer 24

X^2/a^2-y^2/b^2=1

Answer 25

Yes, so to get "a" and "b", we'll need this - http://en.wikipedia.org/wiki/Hyperbola#Quadratic_equation

Answer 26

there's so much work to this section of algebra, it's crazy..

Answer 27

It is, yes. But it is just a little tedious but not hard.

So we were correct that this is a hyperbola and you have the recipe to find a and b. Here is the answer but don't just accept it. Check it using the recipe -

http://www.wolframalpha.com/input/?i=16x%5E2+-81y%5E2+%2B+324+y+%E2%80%93+1620+%3D+0

Answer 28

Why have i never found this website????????

Answer 29

Just use it to check your work - don't let it stop you from learning pls.

$$
a^{2} = -\frac{\Delta}{\lambda_{1}D} = -\frac{\Delta}{\lambda_{1}^{2}\lambda_{2}}\\
b^{2} = -\frac{\Delta}{\lambda_{2}D} = -\frac{\Delta}{\lambda_{1}\lambda_{2}^{2}}
$$
Where all are defined in the last link I sent.

Answer 30

whats that sideways T looking thing mean?

Answer 31

Those are the two roots of the following equation:
$$
\lambda^{2} - \left( A_{xx} + A_{yy} \right)\lambda + D = 0
$$

Answer 32

It's called "Lambda1" and "Lambda2." Lambda \(\Delta\) is a Greek symbol (equivalent to our letter "L")

Answer 33

When you solve this last equation, because it is a quadratic, it will have 2 roots.

Answer 34

Here is the meaning of \(A_{xx}\) and \( A_{yy}\) -
$$
A_{xx} x^{2} + 2 A_{xy} xy + A_{yy} y^{2} + 2 B_{x} x + 2 B_{y} y + C = 0
$$

That's the same equation we had before but with A-F replaced by new names.

Answer 35

For example \( A_{yy}=-81\) from your equation. Make sense?

Answer 36

You're not giving up on me, are you?

Answer 37

Once you do this once, the second time will be easy! Just seems difficult the 1st time.

Answer 38

You were asking good questions.

Answer 39

No sorry i looked off for a moment, was doing something and also reading what you were typing

Answer 40

I think you have everything you need.

Answer 41

So far i understand how to find what type of conic section.
but how about how to find the Verti, Foci, and Center?
or the length of the transverse and conjugate axis?

Answer 42

Center is here
http://en.wikipedia.org/wiki/Hyperbola#Quadratic_equation

Using "Features" (i.e. e,c,l and p) here you can find the other parameters
http://en.wikipedia.org/wiki/Conic_section#Features

You will see how everything ties together here -
http://en.wikipedia.org/wiki/Hyperbola#Nomenclature_and_features

Answer 43

okay ill check it out thanks

Answer 44

send me note if after your review you have questions -- I know you can do this. Show me what you tried

take care

Answer 45

Here's a little animation that might help -


http://assets.openstudy.com/updates/attachments/555fcf87e4b06a1264ab6fc1-ybarrap-1432483249200-hyperbola.gif

Answer 46

Here is where you can interact with my simulation - http://tube.geogebra.org/student/m1235825