SOLUTION: FIND EQUATION OF A PARABOLA WITH FOCAL WIDTH OF 8, AXIS PARALLEL TO Y AXIS, PASSING THROUGH (5,0) AND (9,-6. THANK YOU IN ADVANCE...

Algebra ->  Points-lines-and-rays -> SOLUTION: FIND EQUATION OF A PARABOLA WITH FOCAL WIDTH OF 8, AXIS PARALLEL TO Y AXIS, PASSING THROUGH (5,0) AND (9,-6. THANK YOU IN ADVANCE...      Log On


   

Question 1131569: FIND EQUATION OF A PARABOLA WITH FOCAL WIDTH OF 8, AXIS PARALLEL TO Y AXIS, PASSING THROUGH (5,0) AND (9,-6. THANK YOU IN ADVANCE...
Answer by greenestamps(13334) About Me  (Show Source):
You can put this solution on YOUR website!


If you use this form of the equation for a parabola

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

Then the vertex is (h,k) and the focal width is 4p.

So with the two given points on the parabola, and knowing the focal width is 8, we get two equations in h and k:

(1) 0+=+%281%2F8%29%285-h%29%5E2%2Bk
0+=+%281%2F8%29%28h%5E2-10h%2B25%29%2Bk

(2) 6+=+%281%2F8%29%289-h%29%5E2%2Bk
6+=+%281%2F8%29%28h%5E2-18h%2B81%29%2Bk

Subtracting (1) from (2) eliminates k:

6+=+%281%2F8%29%28-8h%2B56%29
-8h%2B56+=+48
-8h+=+-8
h+=+1

Substituting h=1 in (1) gives us k:

0+=+%281%2F8%29%285-1%29%5E2%2Bk
0+=+2%2Bk
k+=+-2

ANSWER: An equation of the given parabola is y+=+%281%2F8%29%28x-1%29%5E2-2

A graph.... The vertex is (h,k) = (1,-2). p=2 is the distance from the vertex to the focus, so the focus is (1,0); so you can see in the graph that the focal width is 8, with the parabola having x-intercepts -3 and 5.

graph%28400%2C280%2C-10%2C10%2C-4%2C10%2C%281%2F8%29%28x-1%29%5E2-2%29