Hey how on earth do u find the focal length? *embarassed* (1 Viewer)

[MuffinMan]

Helppppppppppppppp

 

[insert-username]

The focal length is equal to a in the equation x² = 4ay. If you're given a question such as this:

x² = 8y. What is the focal length?

You divide the coefficient of y by 4 (from the equation: 8 = 4a, therefore a = 8/4), and then you get a (in this case, a = 2).


If you get a question like this:

A parabola has its vertex at the origin and directrix at y = -5. What is its focal length?

The focal length is equal to the negative of the directrix (in this case, +5), since a parabola is defined as a point that moves so that it is equidistant from a fixed point (focus) and a straight line (directrix). Thus, the focal length is equal to the distance of the directrix from the focus.


If you get this:

A parabola has its vertex at (3,2) and has a directrix of y = 0. What is its focal length?

You subtract the value for the directrix from the y-value of the vertex (i.e. 2 - 0) for the same reason as given above. Here, a = 2.


I hope that helps you,


I_F

 

[MuffinMan]

what if its not in the form x = 4ay???? like 2002 hsc
y = x^2 - 8x + 4
find the coordinates of the focus
sorry, lol i feel so stupid

 

[Trev]

(x-4)²-16=y-4
(x-4)²=y+12
(x-4)²=4*1/4(y+12)
So a=1/4.

 

[chin music]

ye i HATED the focal length for so long. But i went over it a few days ago.
U gotta get the equation into the form, (x-h)^2=4a(y-k). THe focal length is "a" and the vertex is (h,k). And that formula is worth remembering.

 

[ZeGoat]

(x-4)²-16=y-4
(x-4)²=y+10

dude, i'm pretty sure that if you add 16 to -4, you get 12 and not 10

 

[jemsta]

lol
10 chars

 

[Trev]

I hope I don't make stupid mistakes like that tomorrow, or the markers aren't as pedantic as you. (EDITED)