1 ) A highway is to be constructed, passing under the archway of a bridge The archway is in - Maths - Conic Sections

Question

1 ) A highway is to be constructed, passing
under the archway of a bridge The archway is in the form of a half
ellipse 30 m wide and 8 m high Separate lanes, each 3 m wide, are
to be built on the highway such that each lane can allow a truck of
maximum height of 4 m to pass under the bridge
How many lanes can be built on the
highway?

Aakash Sharma · Accepted Answer

Given: Width of the ellipse = 30 m â‡’AA' = 30 m â‡’AC + CA' = 30 m â‡’AC + AC = 30 m (âˆµ AC = A'C) â‡’2AC = 30 m $\Rightarrow AC = \frac{30}{2} m = 15 m$ and height of the half ellipse = 8 m â‡’ BC = 8 m Thus the equation of the given ellipse is $\frac{x^{2}}{(15)^{2}} + \frac{y^{2}}{(8)^{2}} = 1 \Rightarrow \frac{x^{2}}{225} + \frac{y^{2}}{64} = 1$ Since truck of maximum height 4 m can pass hence lanes can only be built in the region having height more than 4 i e NN' Now putting y = 4 we get $\frac{x^{2}}{225} + \frac{4^{2}}{64} = 1 \Rightarrow \frac{x^{2}}{225} + \frac{16}{64} = 1 \Rightarrow \frac{x^{2}}{225} + 1 - \frac{16}{64} = 1 - \frac{1}{4} = \frac{4 - 1}{4} = \frac{3}{4} \Rightarrow x^{2} = \frac{3 \times 225}{4} \Rightarrow x = \sqrt{\frac{225}{4} \times 3} = \frac{15}{2} \sqrt{3} \Rightarrow CN = \frac{15}{2} \sqrt{3}$ â‡’ NN' = CN + CN' = CN + CN (âˆµ CN' = CN) $= 2 CN = 2 \times \frac{15}{2} \sqrt{3} = 15 \sqrt{3} m$ also the width of each lane = 3m $\Rightarrow number of lanes in NN' = \frac{15 \sqrt{3}}{3} = 5 \sqrt{3} = 5 \times 1 732 = 8 66$ Since the number of lanes cannot be in decimals Hence the required number of lanes is 8