How to associate a second degree equation to a bridge arch.

Question

The bridge connects two hills 100 feet apart. The arch on the bridge is in a parabolic form. The highest point on the bridge is 10 feet above the road at the middle of the bridge. Find the equation of the parabola.

My working :

I assumed the middle point of the bridge as (0,0) One end of the bridge is (-50,0) and other end is (50,0) the highest point is (0,10) Let us assume that the equation of the parabola is $$y =-ax^2 +bx + c $$

As this equation satisfy the above three points therefore,

0 = -2500a -50b+c ....(i) [ by putting x = -50, y = 0 ]

0 = -2500a +50b +c ....(ii) [ by putting x = 50, y = 0 ]

10 = c ....(iii) [ by putting x = 0, y = 10 ]

Is this the correct approach , as i am not getting the answer given in the options.

Please guide on this .. thanks

Answer 1

Your approach, leading to the equations (i)–(iii), is correct. It gives $b=0$ (since the bridge is symmetric), $c=10$, and $a={1\over250}$. The equation of the actual parabola then is $$y=-{x^2\over250}+10\ .$$

Answer 2

parabola equation $ x^2 = -4ay$

it is given that height 10 feet is at the middle of the bridge. So a point on parabola is (50,-10). By putting point in above parabola equation, $(50)^2 = -4a*(-10)$

so, $a = 2500/40$

$ a = 250/4$

hence, required equation is : $x^2 = -250y$

have a nice day..........