Question 802814
If it passes through point (6,8), it would need to pass through (0,8); the vertex will be at (3,15), and the base will be from a point (-b,0), to a point (6+b,0).
That makes it complicated.
 
Let's put the origin (0,0) at a point on the ground, directly below the vertex.
We make the line that passes through the feet of the arch the x-axis.
We call the vertical axis of symmetry of the arch our y-axis.
We measure x (the horizontal distance to the origin) in meters,
and y or h(x) (the height above the ground) also in meters.
{{{h(x)=ax^2+15}}} is our equation
The vertex will be (0,15).
The points at a height of 8 meters will be (-3,8) and (3,8), so that the arch width at that height is the (horizontal distance from (-3,8) to 3,8, which is 6.
Substituting the coordinates of (3,8), we get
{{{8= a*3^3+15}}}
{{{8=9a+15}}}
{{{8-15=9a}}}
{{{9a=-7}}}
{{{a=-7/9}}}
{{{highlight(h(x)=(-7/9)x^2+15)}}}
At the base, {{{h(x)=0}}} and
{{{(-7/9)x^2+15=0}}} needs to be solved
{{{(-9/7)(-7/9)x^2=-15(-9/7)}}}
{{{x^2=15*9/7}}}
{{{x^2=135/7}}}
The feet of the arch are at {{{x=-sqrt(135/7)}}} and {{{x=sqrt(135/7)}}}
and the width at the base is
{{{sqrt(135/7)-(-sqrt(135/7))=2sqrt(135/7)}}}= approx. 7.75 meters