Question 1206318
.
The span of the parabolic arch in Figure B is 8 m. 
At a distance of 2 m from the center, the vertical clearance is 4.5 m. 
Find the height of the arch. 
Hint: Choose a convenient coordinate system in which the equation of the parabola 
will have the form x2 4p(y k). Figure B: https://ctrl.vi/i/z9urPS17M
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        I will solve the problem, but will not follow the hint,
        because I have better way to solve.



<pre>
Place the origin of the coordinate system at the ground, where the left branch
of the parabola meets the ground.


Then the parabola has left x-intercept at x= 0 and right x-intercept at x= 8.

So, I can write the parabola equation in the form  

    y(x) = -a*(x-0)*(x-8),  or  y = -ax*(x-8).


Here "a" is some positive real number.  With the sign "-" before "a", the parabola is opened downward.

The parameter "a" is the only unknown now, and now my task is to find its value.


About this parabola, I know that the value y(x) at x = 6 is 4.5; so I write this equation

    y(6) = 4.5.


It is the same as

    -a*6*(6-8) = 4.5,  or  -6a*(-2) = 4.5,  or  12a = 4.5.


From this equation, I find  a = {{{4.5/12}}} = {{{9/24}}} = {{{3/8}}} = 0.375.


So, the parabola equation is y(x) = -0.375x*(x-8).


The maximum height is at x= 4:  {{{y[max]}}} = y(4) = -0.375*4*(4-8) = -0.375*4(-4) = 0.375*16 = 6.


<U>ANSWER</U>.  The maximum height of the arch is 6 meters.
</pre>

Solved.


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When people from outside ask me to solve a problem, I don't like to use 
somebodies' hints - because often I have my own, much better plan/idea/solution in my head.



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&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Word-problems-on-engineering-constructions-of-parabolic-shapes.lesson>Word problems on engineering constructions of parabolic shapes</A> 

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