Parabola Word Problems
Parabola has many applications in our day to day life. For example if an object (projectile) is thrown in space, then the path of the projectile is a parabola.Parabolic reflectors have the property that the light rays or sound waves coming parallel to its axis converge at the focus and then it reflects them parallel to the axis.Due to this property, parabolic reflectors are used in cars, automobiles,loud speakers,solar cookers, telescopes etc.
The various properties of parabola are used to solve parabola word problems.
Lets solve some parabola word problems.
Ex 1 : If a parabolic reflector is 20 cm in diameter and 5 cm deep, find its focus.
Solution : Let LAM be the parabolic reflector such that LM is its diameter and AN is its depth.It is given that AN = 5 cm and LM = 20 cm
`:.` LN = NM = 10 cm.
Taking A as the origin, AX along x- axis and a line through A perpendicular to AX as y-axis, let the equation of the reflector be
y2 = 4ax
The coordinates of L(5,10) lies on the parabola .Therefore,
102 = 4a*5 `=>` 100 = 20a `=>` a = 5
So, the equation of the reflector is
y2 = 20x
Its focus is at (5,0) i.e. at point N.
Ex 2: The focus of a parabolic mirror as shown in figure is at a distance of 6 cm from its vertex. If the mirror is 20 cm deep, find the distance LM.
Solution: Let the axis of the mirror be along the positive direction of x-axis and the vertex A be the origin.
Since the focus is at a distance of 6 cm from the vertex.Then , the coordinates of the focus are ( 6,0).Therefore, the equation of the parabolic section is
y2 = 24x [ Putting a = 6 in y2 = 4ax]
Since L (20,LN) lies on this parabola. Therefore,
LN2 = 24 x 20 `=>` LN = `sqrt(480) = 4xxsqrt(30)`
`:.` LM = 2LN = 8`sqrt(30)` cm
Ex 3 An arc is in the form of a parabola with its axis vertical.The arc is 10m high and 5 m wide at the base. How wide is its 2m from the vertex of the parabola.
Solution: Let the vertex of the parabola be at the origin and axis along OY.Then, the equation of parabola is
x2 = 4ay
The coordinates of end A of the arc are (2.5,10) and it lies on the parabola
`:.` (2.5)2 = 4a * 10
`=>` 6.25 = 40a
`=>` a = ` (6.25)/40 = 625/4000 = 5/32`
Putting the value of "a" in the equation of parabolic arc, we get
x2 =4 * `5/32` * y
`=>` x2 = `5/8` y
When y =2, we have
x2 = `5/8` * 2 = `5/4`
`=>` x = `sqrt(5/4)` =`(sqrt(5))/2`
Hence , the width of the arc at a height of 2m from the vertex is 2 x `(sqrt(5))/2` = `sqrt(5)` m.