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A highway overpass has a parabolic shape. If the edge of the highway is the origin and...
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Let us choose coordinates such that 1 smallest division in both x- and y- directions represent a distance of 1 m. The parabolic overpass touches the highway at its two edges, one of which is the origin, i.e. having coordinates (0,0). Therefore, the origin, is a point on the parabola. The highway is 10 m wide. So, the other edge, i.e. point (10,0) must lie on the parabola. Again the overpass is 13m high at a distance of 2m from the the edge. So, the point (2,13) must also lie on the parabola.
Let us further assume that the equation of the parabola is the quadratic: `y=ax^2+bx+c`
Putting the three points mentioned above, we get
`rArr c=0` --- (i)
`rArr 10a+b=0` --- (ii)
`rArr 4a+2b=13 ` --- (iii)
Solving equations (ii) and (iii) yields
and `b= 130/16`
Therefore, the equation of the parabola takes the form,
This is the required equation of the parabolic overpass.
Check: This is a `x^2=4py` type of parabola, so its axis is vertical. Again, it has p<0, so it must open downwards.
Posted by llltkl on July 23, 2013 at 6:31 PM (Answer #1)
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