Question

An experimental model for a suspension bridge is built in the shape of a parabolic arch. In one section, cable runs from the top of one tower down to the roadway, just touching it there, and up again to the top of a second tower. The towers are both 4 inches tall and stand 40 inches apart. At some point along the road from the lowest point of the cable, the cable is 0.64 inches above the roadway. Find the distance between that point and the base of the nearest tower.

Answer 1

So hmm notice the picture below

based on the info, we can say, is a parabola with that equation

notice, the towers are 40 inches apart, -20 to 20

hmmm ok, so, what's the coefficient "a"?

well, we also know the parabola has a point at 20,4 and -20, 4

so, let's use 20,4

$\bf y=ax^2\qquad \begin{cases} x=20\\ y=4 \end{cases}\implies 4=a20^2\implies \cfrac{4}{400}=a\implies \cfrac{1}{100}=a \\\\\\ thus\implies y=\cfrac{1}{100}x^2\iff y=\cfrac{x^2}{100}$

now... notice in the picture, the red point at (x, 0.64)

so, what's "x" when y = 0.64?

well $\bf 0.64=\cfrac{x^2}{100}\implies 64=x^2\implies \pm\sqrt{64}=x\implies \pm 8 = x$

so.. now we know "x" is 8 or -8, thus the points are 8, 0.64 on the 1st quadrant and -8, 0.64 on the 2nd quadrant

so.. .let's say, let us use the 8, 0.64  point

$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 8}}\quad ,&{{ 0.64}})\quad % (c,d) &({{ 20}}\quad ,&{{ 4}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}$

and surely you'd know what that is