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ind the equation of the parabola with vertex on Ox, parallel to Oy, and passing through (1,-4),(-
1,-1).
analytic geometry
conic sections, the parabola
Solution
You can write the equation of a parabola as:
y = ax^2 + bx + c
Since the vertex is on ox axis, then the coordinates of vertex are (-b/2a, -delta/4a), where delta=0.
delta = b^2-4ac = 0
b^2 = 4ac
b=2sqrt(ac)
The parabola passes through the points (1,-4),(-1,-1):
-4 = a + b + c (1)
-1 = a - b + c (2)
We'll add (1)+(2):
2(a+c) = -5
a+c = -5/2
ac = b^2/4
ac = b^2/4
x^2 + 5x/2 + b^2/4 = 0
delta = 25/4 - b^2
delta = 0
25/4 - b^2 = 0
b^2=25/4
b1=5/2 and b2 = -5/2
a=c=-5/4
The equations of the parabola are:y = -5x^2/4 + 5x/2 - 5/4
and y = -5x^2/4 - 5x/2 - 5/4

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ind the equation of the parabola with vertex on Ox, parallel to Oy, .pdf

  • 1. ind the equation of the parabola with vertex on Ox, parallel to Oy, and passing through (1,-4),(- 1,-1). analytic geometry conic sections, the parabola Solution You can write the equation of a parabola as: y = ax^2 + bx + c Since the vertex is on ox axis, then the coordinates of vertex are (-b/2a, -delta/4a), where delta=0. delta = b^2-4ac = 0 b^2 = 4ac b=2sqrt(ac) The parabola passes through the points (1,-4),(-1,-1): -4 = a + b + c (1) -1 = a - b + c (2) We'll add (1)+(2): 2(a+c) = -5 a+c = -5/2 ac = b^2/4 ac = b^2/4 x^2 + 5x/2 + b^2/4 = 0 delta = 25/4 - b^2 delta = 0 25/4 - b^2 = 0 b^2=25/4 b1=5/2 and b2 = -5/2 a=c=-5/4 The equations of the parabola are:y = -5x^2/4 + 5x/2 - 5/4 and y = -5x^2/4 - 5x/2 - 5/4