SOLUTION: Find the equation of the parabola with vertex on the line y = 2, axis parallel to Oy, latus rectum 6, and passing through (2, 8).

Algebra ->  Finance -> SOLUTION: Find the equation of the parabola with vertex on the line y = 2, axis parallel to Oy, latus rectum 6, and passing through (2, 8).      Log On


   

Question 1190798: Find the equation of the parabola with vertex on the line y = 2, axis parallel to Oy, latus rectum 6, and passing through (2, 8).
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
the equation of the parabola with latus rectum +6
length of latus rectum is +6, so the leading coefficient is ±1%2F6
vertex on the line y+=+2=>k=2
so far we have:
y+=+%281%2F6%29%28x-h%29%5E2%2B2
passing through (2, 8)
use it to calculate h


8+=+%281%2F6%29%282-h%29%5E2%2B2
8+=+h%5E2%2F6+-+%282+h%29%2F3+%2B+8%2F3 .......multiply by common denominator 6
48+=+h%5E2+-+4+h+%2B+16
+h%5E2+-+4+h+%2B+16-48=0
+h%5E2+-+4+h+-32=0
%28h+%2B+4%29+%28h+-+8%29=0
=>h=-4 or h=8

and, your equation is:
y+=+%281%2F6%29%28x%2B4%29%5E2%2B2
or
y+=+%281%2F6%29%28x-8%29%5E2%2B2

graph: y+=+%281%2F6%29%28x%2B4%29%5E2%2B2


graph: y+=+%281%2F6%29%28x-8%29%5E2%2B2