4 Finding the equation of a parabola

Consider a parabola that has its vertex at s = 50 when t = 0 and rises to s = 100 when t = 30 . In coordinate terms, we need the equation of a parabola that has its lowest point or vertex at (0, 50) and passes through (30, 100). The general form

y − C = A ( x − B ) 2

is useful here.

In this case y corresponds to s and x to t . So the equation relating s and t is

s − C = A ( t − B ) 2

According to the general form, the coordinates of the vertex are ( B , C ) . We know that the coordinates of the vertex are (0, 50). So we can deduce that B = 0 and C = 50 . It remains to find A . The fact that the parabola must pass through (30, 100) may be used for this purpose. These values together with those for B and C may be substituted in the general equation:

100 − 50 = A ( 30 − 0 ) 2

so 50 = 900 A or A = 1 18 and the function we want is

s = 50 + 1 18 t 2 ( 0 ≤ t ≤ 30 )

Task!

Find the equation of a parabola with vertex at ( 0 , 2 ) and passing through the point ( 4 , 4 ) .

Using the general form, with B = 0 and C = 2 ,

y − 2 = A ( x − 0 ) 2 or y − 2 = A x 2

Then using the point ( 4 , 4 )

4 − 2 = 16 A so A = 2 16 = 1 8

and the required equation is

y = 2 + 1 8 x 2

Exercise

An open-topped carton is constructed from a 200 mm × 300 mm sheet of cardboard, using simple folds as shown in the diagram.

{Cardboard folds to make an open-topped carton}

  1. Show that the volume of the carton (in cm 3 ) is

    V = x ( 300 − 2 x ) ( 200 − 2 x ) 1000

    so V = x 3 250 − x 2 + 60 x …(*)

  2. Sketch Equation (1) as V vs x and hence estimate the maximum volume of carton that may be obtained by folding the cardboard sheet.
  3. A carton with a volume of 1000 cm 3 is to be made from the cardboard sheet.
    1. Show that one solution is to use a height x = 50 mm.
    2. By factorisation of Equation (*) for V = 1000 cm 3 , find a second solution for x which would give the same carton volume.
    3. Why does the third root have no physical meaning?
  1. V = x ( 300 − 2 x ) ( 200 − 2 x ) 1000 = x 3 250 − x 2 + 60 x ( cm 3 )
  2. No alt text was set. Please request alt text from the person who provided you with this resource. V max ≈ 1056 cm 3 when x ≈ 39.2
    1. x = 50 mm    ⇒    V = 1000 cm 3 as required.
    2. x 3 − 250 x 2 + 15000 x 250 − 1000 = 0 factorises to

      ( x − 50 ) ( x 2 − 200 x + 5000 ) = 0

      so x = 50 or x = 100 ± 10 50 ≈ 29.3 or 170.7 . The second root is 29.3 .

    3. The third root 170.7 is impossible as 200 − 2 x must be a positive distance.