Practical Problems Using Parabola Ellipse and Hyperbola :
Here we are going to see some practice problems on parabola, ellipse and hyperbola.
Question 1 :
A bridge has a parabolic arch that is 10 m high in the centre and 30 m wide at the bottom. Find the height of the arch 6 m from the centre, on either sides.
Solution :
From the given information, the parabola is symmetric about y-axis and it is open down word.
Since the vertex is (h, k),
(x - h)2 = -4a (y - k)
Here (h, k) ==> (0, 10)
(x - 0)2 = -4a (y - 10)
x2 = -4a (y - 10) ----(1)
The parabola is passing through the point (15, 0).
152 = -4a (0 - 10)
225/40 = a
a = 45/8
By applying the value of a in (1), we get
x2 = -4 (45/8) (y - 10)
x2 = -(45/2) (y - 10)
By applying x = 6 and y = h, we get
62 = -(45/2) (h - 10)
-36(2/45) = h - 10
(-72/45) + 10 = h
h = -1.6 + 10
h = 8.4 m
Hence the required height is 8.4 m.
Question 2 :
A tunnel through a mountain for a four lane highway is to have a elliptical opening. The total width of the highway (not the opening) is to be 16 m, and the height at the edge of the road must be sufficient for a truck 4 m high to clear if the highest point of the opening is to be 5 m approximately . How wide must the opening be?
Solution :
(x2/a2) + (y2/b2) = 1
The elliptical shape is passing through the point (8, 4).
(82/a2) + (42/b2) = 1
Here b = 5
(82/a2) + (42/52) = 1
(82/a2) = 1 - (16/25)
(64/a2) = (25 - 16)/25
(64/a2) = 9/25
a2 = 64 (25/9)
a = 8(5)/3
a = 40/3
2a = 80/3
2a = 26.6
Hence the opening must be 26.6 m
Let us look into the next example on "Practical Problems Using Parabola Ellipse and Hyperbola".
Question 3 :
At a water fountain, water attains a maximum height of 4 m at horizontal distance of 0 5 . m from its origin. If the path of water is a parabola, find the height of water at a horizontal distance of 0.75 m from the point of origin.
Solution :
The parabola is symmetric about y-axis and it is open downward.
(x - h)2 = -4a(y - k)
Te vertex of the parabola is (0.5, 4)
(x - 0.5)2 = -4a(y - 4) ----(1)
The parabola is passing through the point (0, 0).
(0 - 0.5)2 = -4a(0 - 4)
0.25 = 16a
a = 0.25/16
a = 1/64
By applying the value of a in (1), we get
(x - 0.5)2 = -4(1/64)(y - 4)
(x - 0.5)2 = (-1/16)(y - 4)
The parabola is passing through the point (0.75, h), we get
(0.75 - 0.5)2 = (-1/16)(h - 4)
(0.25)2 = (-1/16)(h - 4)
-0.0625(16) = h - 4
-1 + 4 = h
h = 3
Hence the required height is 3 m.
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