Hyperbolas and Parabolas: find focus, etc.

[mike_vidov]

Hello, I am currently behind in my class and need assistance to catch up with the class. Can anyone do and explain these two problems so I know.

Hyperbolas: Find the center, vertices, foci, and asymptotes. Then graph.

. . . . .\(\displaystyle \L{\frac{(x\,-\,2)^2}{9}\,-\,\frac{(y\,+\,5)^2}{1}\,=\,1}\)

Parabolas: For each parabola with vertex at the origin, find the focus and directrix.

. . . . .\(\displaystyle \L{x^2\,=\,16y}\)

Thanks, I really appreciate it.

 

[tkhunny]

These are already in a nice standard form. Just compare the structure to the notes in your text and you're done.

 

[ktydecou]

Here are some more hint in case you need the the standard form is: (sorry I don't know how to write it on a computer, but I will do my best)
(x-h)^2 - (y-k)^2=1
--------- ---------
a^2 b^2

You take the square root of both a^2 and b^2 so, a=3 and b=1
So the center would be (h,k). In your problem (2, -5)
Then the vertices are one below and above your center. (2, -4) and (2,-6)
Then dram a rectangle around your center that is 2a=6 units high and 2b= 2long
Then draw your asympototaes through the coners of your box
Then draw your hyperbola from your vertices following your asympototes
I hope that helps you if you don't already have the answer!

For the parabola the center is at (0,0) so divide 16 by 4 becasue standard form is:
x^2=4py (for a vertical positive porabola)
Then you get 4 for your answer and that is your focus (0,4) adn the directrix is
y= -p so y= -4
Remeber that different porabolas have different standared equations
I hope this help
BYE!

 

[mike_vidov]

Thanks man, helps a ton. I really envy all of you on here who can remember and do these like this. Its inspiring. Thanks again.