Question

PLEASE ASAP
Find the vertex, focus, directrix, and focal width of the parabola.

-(1/12)x^2 = y

Vertex: (0, 0); Focus: (0, -3); Directrix: y = 3; Focal width: 48

Vertex: (0, 0); Focus: (-6, 0); Directrix: x = 3; Focal width: 48

Vertex: (0, 0); Focus: (0, 3); Directrix: y = -3; Focal width: 3

Vertex: (0, 0); Focus: (0, -3); Directrix: y = 3; Focal width: 12

Answer 1

Well, the equation is of the form x^2 = 4py. Notice how I put x^2 by itself, though. That's one tip to start off with : )

Answer 2

so is it A?

Answer 3

Focal width is equal to 4p. I know the temptation is to multiply that -12 by 4, but it already is 4p. So that 12 is your focal width.

Answer 4

ohhhhhh okay I understand. What about this one: Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ±1 divided by 4.x.

y squared over 16 minus x squared over 64 = 1

y squared over 16 minus x squared over 256 = 1

y squared over 256 minus x squared over 16 = 1

y squared over 64 minus x squared over 4 = 1 ?????

Answer 5

Well, since the y-coordinates of the verticies are changing, its a y^2/a^2 form where a = the distance from the center to the vertex. So the distance is 4, meaning a^2 = 16. as for the asymptotes, when the hyperbola is in y^2/a^2 form, the asymptotes are at +/- (ax/b). So if a, which is your numerator, is 4, then what does b have to be in order for it to come out to 1/4?

Answer 6

16?

Answer 7

There ya go. So if b is 16, b^2 has to be 256.

Answer 8

oh gosh, ty so much

Answer 9

You should maybe keep track of some of these formulas and equations, lol.

Answer 10

I have them all with me now, I just didn't understand how to start the problem.

Answer 11

Ah. Well,it just gives you basic info. With hyperbolas, the verticies always give you a. Which coordinate changes between the verticies is what tells you y^2 first or x^2 first. You can derive everything from certain basic things given in the problem :P