SOLUTION: find the equation of a parabola with vertex on the line y=2x, axis parallel to the x-axis and passing through (3/2, 1) and (3,4).

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: find the equation of a parabola with vertex on the line y=2x, axis parallel to the x-axis and passing through (3/2, 1) and (3,4).       Log On


   

Question 1187647: find the equation of a parabola with vertex on the line y=2x, axis parallel to the x-axis and passing through (3/2, 1) and (3,4).


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The given information is unusual, so initially we don't have a good idea of where to go with the problem. So we do the only thing we can -- write the general form of the equation and use each the two given points in that equation to see what it gives us.

The vertex is on the line y=2x, so we can call the coordinates of the vertex (a,2a). Then the vertex form of the equation of a parabola parallel to the x-axis with vertex (a,2a) is

%28x-a%29+=+%281%2F%284p%29%29%28y-2a%29%5E2

Plug in (x,y)=(1.5,1) and (x,y)=(3,4) to get two equations in a and p:

1.5-a=%281%2F%284p%29%29%281-2a%29%5E2

4p%281.5-a%29=1-4a%2B4a%5E2 [1]

3-a=%281%2F%284p%29%29%284-2a%29%5E2

4p%283-a%29=16-16a%2B4a%5E2 [2]

Subtract [1] from [2]:

4p%281.5%29=15-12a
6p=15-12a
p=2.5-2a
4p=10-8a [3]

Substitute [3] in [2]:

%2810-8a%29%283-a%29=16-16a%2B4a%5E2
30-34a%2B8a%5E2=16-16a%2B4a%5E2
4a%5E2-18a%2B14=0
2a%5E2-9a%2B7=0
%282a-7%29%28a-1%29=0

a=3.5 or a=1

There will be two parabolas that satisfy the given conditions.

(1) a=3.5; 4p=10-8a=-18; 2a=7

x-3.5=%28-1%2F18%29%28y-7%29%5E2

(2) a=1; 4p=10-8a=2; 2a=2

x-1=%281%2F2%29%28y-2%29%5E2

Here is a graph of the two parabolas, one with vertex (1,2) and the other with vertex (3.5,7) and both passing through the points (1.5,1) and (3,4)