The height of the arch from the ground will be 125 yards.
Given information:
An architect is designing a commemorative steel arch in the shape of a parabola.
Parabola is vertically placed and its legs are 300 yards apart.
The focus of the parabola is 80 yards above the ground.
Let a be the distance of focus from the vertex of the parabolic arch.
So, the equation of parabola will be,
[tex]x^2=-4ay[/tex]
Assume the origin is placed on the vertex of the parabola.
So, the coordinates of the legs will be,
(-150, -(80+a)) and (150, -(80+a))
Satisfy the above point with the equation of parabola as,
[tex]x^2=-4ay\\150^2=4a(80+a)\\a^2+80a-5625=0[/tex]
Solve the above quadratic equation as,
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\a=\dfrac{-80\pm\sqrt{80^2+4\times 1\times5625}}{2}\\a=\dfrac{-80\pm170}{2}\\a=-125, 45[/tex]
The value of a can't be negative. So, [tex]a=45[/tex].
So, the height of arch will be,
[tex]80+a=80+45\\=125\rm\;yd[/tex]
Therefore, the height of the arch from the ground will be 125 yards.
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