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Binomial Theorem

Question
CBSEENMA11015136

An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2m from the vertex of the parabola? 

Solution

Let AB be the parabolic arch having O at the vertex and the vertical line OY as the axis.
The parabola open upwards
∴   Its equation is of the form WiredFaculty ...(i)
     Width of the arch, LM = 5 m
       rightwards double arrow                    OM = 2.5 m
      Height of the arch, BM = 10 m
∴ Co-ordinates of point B are (2.5, 10)
Since point B lies on the parabola WiredFaculty
∴     left parenthesis 2.5 right parenthesis squared space equals space 4 straight a left parenthesis 10 right parenthesis space rightwards double arrow space 25 over 10 cross times 25 over 10 space equals space 40 straight a rightwards double arrow space straight a space equals space 25 over 160 space equals space 5 over 32
∴     From (i), the equation of the parabola is: WiredFaculty
or                       straight x squared equals 5 over 8 straight y                                                           ...(ii)
We have to find the width PQ of the arch at a distance ON = 2 m from the vertex.
Let                       PQ = d rightwards double arrow NQ = WiredFaculty
∴    Co-ordinates of point Q are open parentheses straight d over 2 comma space 2 close parentheses
Putting it in (ii), we get straight d squared over 4 space equals space 5 over 8 left parenthesis 2 right parenthesis space rightwards double arrow space straight d squared space equals space 5
Hence, the width of the arch = d = square root of 5 space straight m or 2.23 (approx).
                                                    WiredFaculty

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