Finding the Width at the Bottom of a Vertical Parabolic Arc

For a National Board Exam Review:

An arc 18m high has the form of a parabola with the axis vertical. If the width of the arc 8m from the top is 64m, Find the width of the arc at the bottom.

Construct Equation:

$${ (y-k) = -4a(x-h)^2 }$$

Assume ${(h,k) = (0,18) }$ and ${(x,y) = (64/2,10)}$

$${ (10-18) = -4a(32)^2 ; a = \frac{-1}{152} }$$

$${ a = \frac{1}{152} }$$

${ Change (x,y) to (x,0) }{ (0-18) = -4(\frac{1}{152})(x-0)^2 ; x=26.15 ... }$$What am I doing wrong? Any Hint? • where you wrote 18-10 should be 10-18 – Chester Aug 20 '15 at 0:54 • @Chester edited! – james Aug 20 '15 at 1:03 • ... which would only swap a sign, but your a is incorrect as well. FYI: it is quicker to do the arithmetic for this problem if you write things in powers of 2, i.e. 8=2^3, 4=2^2, 32 = 2^5. – Chester Aug 20 '15 at 1:05 2 Answers Let w > 0 be the desired width. Place the origin at the vertex so that the parabola has the form y = ax^2. Then the parabola goes through the points (32, -8) and (w/2, -18). Plugging in the first point yields:$$ -8 = a(32)^2 \iff a = \frac{-1}{4 \cdot 32} $$Plugging in the second point yields:$$ -18 = \frac{-1}{4 \cdot 32} \cdot \frac{w^2}{4} \iff w^2 = (4 \cdot 4)(2 \cdot 9)(2 \cdot 16) \iff w = 4 \cdot 2 \cdot 3 \cdot 4 = 96 $$Supposing a=1, your horizontal movement when you are 8 metres below the arc's maximum should be its square root, 2 \sqrt2. However, the width of one half the arc is 32, meaning that the arc is \frac {32} {2 \sqrt2} , or 8 \sqrt2 times wider than if a=1. Similarly, at 18 metres below the arc's maximum, the width of one half the arc should be its square root, or 3 \sqrt2. However, we know the arc is 8 \sqrt2 times wider than this, so we multiply the two.$$8 \sqrt2 * 3 \sqrt2 = 24 * 2 = 48. $$Remember, this width is only one half of the arc at its base. Thus,$$48 * 2 = 96$\$

Hope this helped.