Question

Here AB is the longest wire (30m) and (OC) is the shortest at ( 6 mathrm{m} ). LN is the supporting wire, ( 18 mathrm{m} ) from the
middle.
We can see that ( B C= ) half of ( A B=50 mathrm{m} )
The shape of the parabola opens upwards, therefore the equation is ( x^{2}=4 a y )
We can see that coordinates of A is (50,24) (i.e, ( y ) coordinate is ( A B-O C=30-6=24 ) )
since ( A ) is a point on the parabola, ( 50^{2}=4 times a times 24^{2} rightarrow a=frac{625}{24} )
Therefore, the equation of the parabola is ( x^{2}=4 times frac{625}{24} y rightarrow 6 x^{2}=625 y )
( x ) coordinate at point ( L=18 rightarrow 6 times 18^{2}=625 y rightarrow y ) coordinate at point ( L=frac{6 times 18^{2}}{625} approx 3.11 )
We need to calculate LN, which is ( 3.11+6=9.11 mathrm{m} ).

# 1 3. parabola. al Wites The cable of a uniformly loaded suspension bridge hangs in the form of a The roadway which is horizontal and 100 m long is supported by vertical attached to the cable, the longest wire being 30 m and the shortest being Find the length of a supporting wire attached to the roadway 18 m from middle. 1g 6 m.

Solution