Question

The equation of the parabola is of the form ( x^{2}=4 a y ) (as it is opening upwards).
It can be clearly seen that the parabola passes through point ( left(frac{5}{2}, 10right) )
( left(frac{5}{2}right)^{2}=4 a(10) )
( Rightarrow a=frac{25}{4 times 4 times 10}=frac{5}{32} )
Therefore, the arch is in the form of a parabola whose equation is ( x^{2}=frac{5}{8} y ).
When ( y=2 mathrm{m}, x^{2}=frac{2}{8} times 2 )
( Rightarrow x^{2}=frac{5}{4} )
( Rightarrow x=sqrt{frac{5}{4}} mathrm{m} )
( therefore mathrm{AB}=2 times sqrt{frac{5}{4}} mathrm{m}=2 times 1.118 mathrm{m}(text { approx. })=2.23 mathrm{m}(text { approx. }) )
Hence, when the arch is ( 2 mathrm{m} ) from the vertex of the parabola, its width is approximately 2.23
( mathrm{m} )

# I Tucus. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola? T heofiformer ledad en

Solution