find the equation of a parabola with axis parallel to y passing through (-1,-3), (1,-2), (2,1)
The equation of a parabola with a vertical axis is:
(y – k) = a.(x – h)^2
The three points coincident with the curve are,
(-1, -3), (1, -2), (2, 1).
These coordinate points satisfy the equation of the parabola.
Substituting for these coordinate values into the eqn of the parabola gives three simultaneous equations.
(-3 – k) = a.(-1 – h)^2
(-2 – k) = a.(1 – h)^2
(1 – k) = a.(2 – h)^2
Expanding the squared terms,
-3 – k = a(1 + 2h + h^2) ------------- (1)
-2 – k = a(1 – 2h + h^2) ------------- (2)
1 – k = a(4 – 4h + h^2) ------------- (3)
(1) - (2) gives
-1 = 4ah -> ah = -1/4
(2) - (3) gives
-3 = a(-3 + 2h)
-3 = -3a + 2ah
-3 = -3a – 2/4 using ah = -1/4.
3a = 2 ½ = 5/2
a = 5/6
Therefore h = -1/(4a) = -1/(20/6) = -6/20 = -3/10
h = -3/10
substituting for a = 5/6 and h = -3/10 into (1 – k) = a.(2 – h)^2,
(1 – k) = 5/6.(2 + 3/10)^2
1 – k = 5/6.(23/10)^2
1 – k = 5*23^2/(6*100) = 2645/600 = 529/120
k = 1 – 289/120 = -409/120
k = -409/120
The unknown parameters are: h = -3/10, k = -409/120, a = 5/6
The parabola is: (y + 409/120) = 5/6.(x + 3/10)^2, or
The equation of the parabola is: y(x) = (5/6)x^2 + (1/2)x – 10/3