Question

A building has an entry the shape of a parabolic arch 84 ft high and 42 ft wide at the base as shown below. Find an equation for the parabola if the vertex is put at the origin of the coordinate system.

Answer 1

"as shown below" implies that a picture is given. Please post it.

Answer 2

are you like supposed to solve it like this
21^2=4p(-84)
441/-84=4p
-5.25=4p or x^2=-5.3y

Answer 3

okay...I got that part

Answer 4

We know it is a parabola so try using this formula

y = a(x - h)^2 + k

You already know the vertex (0,0) and two points on the parabola (21, -84), (-21,-84)

Answer 5

Sorry, sign errors

Answer 6

You need to make two equations, one for each given point, then solve for a

Answer 7

-84 = a(21 - 0)^2 + 0
-84 = a(-21 - 0)^2 + 0

-84 = a(21)^2
-84 = a(-21)^2

-84 = 441a
-84/441 = a

Answer 8

-4/21 = a

Answer 9

my answers are supposed to be like this
y2 = -21x

x2 = -21y

x2 = -5.3y

y2 = -5.3x

Answer 10

Hmmm

Answer 11

I thought that because the parabola opened down it was supposed to look like x^2= somthing

Answer 12

It is

\[y = -\frac{4}{21}x^2\]

Answer 13

That's the parabola I got

Answer 14

If you graph that, you will see that (-21, -84) and (21, 84) are points on it

Answer 15

I think I see. We have to put the parabola in a different form that matches one of the answer choices.

Answer 16

Try isolating x^2

Answer 17

yeah that's what I just did

Answer 18

but, why did you use that formula ?

Answer 19

never mind ...would you mind helping me with another problem ?Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at .

Answer 20

Because that formula is the equation for a parabola when you know at least one point on the parabola and the vertex. It allows you to plug in two points into one equation.

Answer 21

Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at .

Answer 22

Use the formula for equation of a hyperbola to help you

Answer 23

okay I tried doing this
a=1*2=2^2=4
b=2*2=4^2=16
y^2/64-x^2/16

Answer 24

is that right?

Answer 25

Oh, sorry, I didn't see that you posted a response until now.