Conic word problems

Hi dijasenoj,

Perhaps this will put you on the right track for the first one.

The formula of a parabola is y = ax^2 + bx +c.

Now, the trick in solving the doorway problem is to select how you impose your coordinate system on that doorway, which is 12 feet high and 8 feet wide.

If I decide that the top of the parabola is at origo, I can simplify the problem quite a bit!

First of all, if the top is at origo, both the x and the y coordinate is 0.

0 = a0^2 + b0 + c means c has to be 0.

Moreover, if the top is at origo, then the symmetry axis is along the y axis. And when the symmetry axis is along the y axis, then b = 0.

(The formula for the symmetry axis is x = -b/(2a), which, when x =0 requires b to be 0, but I like to keep the image of the parabola firmly in mind,
because it helps me see that b has to be 0 without the formula for the symmetry axis.)

In other words, the formula has been simplified to

y= ax^2

To find a you can use the coordinates for the "heels" of the portal. The height is 12 feet, so the y-coordinate is -12. The width is 8 feet, and as the symmetry line at x =0 bisects the parabola, the x coordinates are -4 and 4.

When you have a, you can figure out the y-coordinate which corresponds to an x value of 3 feet from the center, (x = 3 or x = -3).

The y-value won't correspond to the height at 3 feet away from the center, you've got to remember where your start point is (at -12 feet). But I think you could work it out from there.

I'm no math specialist, and I've seen other people solve the same kind of problem putting the "heels" of the parabola at height 0 and the top point at height 12, but personally I prefer to have the origo where the derivative is 0.

Thanks Inger Hohler