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(A) The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola,while its vertical axis

Vedclass · Accepted Answer

$9.11$

Vedclass · Answer

(A) The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola,while its vertical axis is taken along the positive $y-$ axis.Here,$AB$ and $OC$ are the longest and the shortest wires,respectively,attached to the cable. $DF$ is the supporting wire attached to the roadway,$18 \, m$ from the middle.Here,$AB = 30 \, m$,$OC = 6 \, m$,and $BC = \frac{100}{2} = 50 \, m$.The equation of the parabola is of the form $x^{2} = 4ay$ (as it is opening upwards).The coordinates of point $A$ are $(50, 30-6) = (50, 24)$.Since $A(50, 24)$ is a point on the parabola,$(50)^{2} = 4a(24)$.$\Rightarrow a = \frac{50 	imes 50}{4 	imes 24} = \frac{625}{24}$.$	herefore$ Equation of the parabola is $x^{2} = 4 	imes \frac{625}{24} 	imes y$,which simplifies to $6x^{2} = 625y$.The $x-$ coordinate of point $D$ is $18$.Hence,at $x = 18$,$6(18)^{2} = 625y$.$\Rightarrow y = \frac{6 	imes 324}{625} = \frac{1944}{625} = 3.1104 \, m$.$	herefore DE = 3.11 \, m$ (approx).$DF = DE + EF = 3.11 \, m + 6 \, m = 9.11 \, m$.Thus,the length of the supporting wire attached to the roadway $18 \, m$ from the middle is approximately $9.11 \, m$.

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