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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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islnds | Valedictorian

Posted July 23, 2013 at 5:01 PM via web

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A highway overpass has a parabolic shape. If the edge of the highway is the origin and the highway is 10m wide, what is the equation of the parabola if the height of the overpass 2m from the edge of the highway is 13m?

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llltkl | College Teacher | Valedictorian

Posted July 23, 2013 at 6:31 PM (Answer #1)

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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

`0=a*0+b*0+c`

`rArr c=0` --- (i)

`0=a*10^2+b*10+0`

`rArr 10a+b=0` --- (ii)

and `13=a*2^2+b*2+0`

`rArr 4a+2b=13 ` --- (iii)

Solving equations (ii) and (iii) yields

`a=-13/16`

and `b= 130/16`

Therefore, the equation of the parabola takes the form,

`y=-13/16x^2+130/16x`

`rArr -13x^2+130x=16y`

`rArr -13(x^2-10x+25)=16y-13*25`

`rArr (x-5)^2=4*(-4/13)*(y-325/16)`

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.

 

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